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COURSE CODE: MTS 105 LECTURE NOTE
COURSE TITLE: ALGEBRA FOR BIOLOGICAL SCIENCES
NUMBER OF UNIT: 03
LECTURER: DR. I. O. ABIALA COURSE OUTLINE: Elementary set theory: set notations, set operations, algebra of sets, Venn
diagram and Applications. Operations with real numbers: indices, Logarithms, surds, uses of Logarithms
in agricultural sciences. Remainder and factor theorems. Partial fractions, Linear and Quadratic
inequalities. Theory of quadratic equations, Cubic equations, equations reducible to quadratic type.
Sequences and series: Arithmetic and Geometric progression, Arithmetic and Geometric Mean,
Arithmetic and Geometric series, nth term of a series, Binomial theorem, Binomial series, the general
term of Binomial series. Matrix algebra: Matrices, algebra of matrices, Determinant of a matrix,
properties and inverse of a matrix, solution of a linear system of equations. Elementary trigonometry:
Degree and Radian measures, Pythagorean identities, Trigonometric functions of any angle, graphs,
inverse trigonometric functions, compound angles and solution of trigonometric equations.
REFERENCES
(1) E.Egbe, G.A. Odili and O.O. Ugbebor: Further Mathematics published by
Africana- FEB Publishers, Gbagada, Lagos, 2000.
SET THEORY
Definition: A set is a collection of objects or things that is well defined.
Here are some examples of sets:
1. A collection of students in form one
2. Letters of the alphabet
3. The numbers 2, 3, 5, 7, and 11
4. A collection of all positive numbers
5. The content of a lady’s purse
The concept of set is very important because set is now used as the official mathematical
language. A good knowledge of the concept of set is, therefore, necessary if mathematics
is to be meaningful to its users.
Notation
A set is usually denoted by capital letters; while the objects comprising the set are written
with small letters. These objects are called members or elements of a set.
For example set A has members .,,, dcba
Convention
The listing of a set A as ,,,, dcba as seen above is not an acceptable mathematical
specification of a set. The correct representation of a set that is listed is to write the
elements, separated by commas and enclosed between braces or curly brackets.
e.g., set { }.,,, dcbaA = .
The statement b is an element or member of set A or b belongs to a′ is written in the
manner .Ab∈ The contrary statement that b does not belong to A is written as: .Ab∉
There are two ways of specifying a set. One way is by listing the elements in the set,
such as:
{ }.,,, dcbaA = .
A second way of specifying a set is by stating the rule or property which characterizes the
set.
For example, { }.52/ <<= xxB or { }.52/ <<= xxB . Notice, the stroke/or colon: can be
used interchangeably, with each as ‘such that’. The representation, { }.52/ <<= xxB is
read as follows:
B is a set consider of elements ,x such that 2 is less than x and x is less than 5.
If a set is specified by listing its elements, we call it the tabular form of a set; and if it is
specified by stating its property, such as { xxC /= is odd }, then it is called the set builder
form.
Finite and Infinite Sets
A finite set is one whose members are countable: for example, the set of students in
Form 1. Other examples are:
(i) the contents of a lady’s hand-bag;
(ii) whole numbers lying between 1 and 10;
(iii) members of a football team.
The finite set is itself in exhaustive; readers can give other examples of a finite set.
An infinite set is one whose elements are uncountable, as they are infinitely numerous.
Here are a few examples of the infinite sets;
(i) Real numbers.
(ii) Rational numbers
(iii) Positive even numbers
(iv) Complex numbers
The main distinction between a finite set and that a finite set has a definite beginning and
a definite end, while the infinite set may have a beginning and no end or vice versa or
may not have both beginning and end.
For example, we specify the set of positive even numbers, as follows:
{ },...6,4,2=P or
{ xxxP ,2: >= is even }
The set of real whole numbers which end with the number 3 is written as follows:
SUBSETS
Suppose { }fedcbaP ,,,,,= and { }.,, edcQ = , then we say Q is contained in ,P and we
use symbol '' ⊂ to denote the statement ‘is contained in’, or ‘is subset of’. Thus PQ ⊂ , is
ready as Q' is contained in 'P . More aptly put, Q is contained in P if there is an ,x such
that Qx ∈ implies .Qx ∈ The statement Q is contained in P can be put in reverse order as
P' contains 'Q and we write .QP ⊃ However, this form is not very popular. If Q is not a
subset of set { },,4,3 aR = then we write RQ ⊄ . It should be noted that unless every
member or Q is also a member of P , then can we say Q is subset of .P
EQUITY OF SETS
Two sets of X and Y are equal if and only if YX ⊂ and XY ⊂ . Suppose { }3,2,1=X and
{ }2,1,3=Y and YX = . Note that the rearrangement of the elements if a set does not alter
the set.
TYPES OF SETS
Null or Empty Sets
Null means void, therefore, a null set is an empty set, or a set that has no members. The
null set is denoted by the symbol { } . Note that { }0 cannot be classified as a null set,
because it has an element, zero.
Singleton
Any set which has only one member is called a singleton. e.g., ( )a is a singleton.
The Universal Set
Set is a subset of a larger set is called the universal set or empty, the Universe of
Discourse.
Thus, in any given context, the total collection of elements under discussion is called the
Universal set.
The symbol U or Ε is often used to denote a universal set. For example, if we toss a die,
once, we expect to have either ,5,4,3,2,1 or ,6 as an end result. If there are no other
expected results different from this numbers, then we say, for this particular experiment,
the universal set is { }6,5,4,3,2,1 . Thus a universal set is the total population under
discussion.
Proper Subsets
If P is a subset of Q and if there is at least one member of Q which is not a member of P ,
then P is a proper subset of Q and we write .QP ⊂
Consider the set =A { }.,3,2,1 The following sets
{ } { } { } { } { }{ } { } { },,3,,2,1,,3,2,,3,,1,,2,1,,3,2,1 are subsets of A . The set { }3,2,1 is
not a proper subset of A ; whereas all others including { }are proper subsets of A . Thus
{ }3,2,1 ⊄ { },3,2,1 but { }3,2,1 , { }3,2,1 ⊆ { }.3,2,1
Power Set
The collection of all the subsets of any set S is called the power set of S . If a set has
n members, where n is finite, then the total number of subsets of S is .2 n Occasionally
we denote the power set S by .2 S
For example: Let { }cbaA ,,= . The subsets of A are { }cba ,, ,{ }ba, ,{ }ca, ,{ }cb, , { }a ,{ }b ,{ }c and
{ }. The power set of A written 32)( =Ap subsets; as seen above.
Example: Find the power set S2 of the sets
(a) { )4,3=S
(b) { { } }2,1,,aS =
Solution:
(a) { } { } { } { },,4,3,4,3=S
(b) { { } } { } { } { },2,1,,2,1,,2 aaP =
In this example, (b) contains only two elements a and { }2,1
Venn – Euler Diagrams
The theory of set can be better understood if we make use of the Venn-Euler diagrams.
The Venn-euler diagram is an instructive illustration which depicts relationship between
sets.
Suppose YX ⊂ and YX ⊄ , we can represent this statement in a Venn-Euler diagram as
follows:
Y Y
X X
X
A B
Set Operations
In set, we use the symbols ∪ read ‘unions and ∩ read ‘intersection’ as operations. These
operations are similar but not exactly the same as the operations in arithmetic. At the end
of this chapter, a reader of this topic should be able to identity areas of analogy between
the operation in arithmetic and those of a set.
Union of Sets
Definition
The union of sets A and B is the set of all elements which belongs to A or B or to both
A and B . This is usually written as BA ∪ , and read A' union '.B
In set language, we define BA ∪ as:
{ AxxBA ∈=∪ : or }B∈ .
The shaded portions in the Venn-Euler diagram in BA ∪
A B A B
The Intersection of Sets
The intersection of sets AandB is the set of elements which belong to both AandB .
Simply, A' intersection 'B written BA ∩ consists of elements which are common to both
AandB .
The Venn-Euler diagram which represent BA ∩ is shaded portion.
In set language
{ BAandxxxBA ∈∈=∩ :
Complement of Sets
The complement of a set x is the set of elements which do not belong to x , but belong to
the universal set. The complement of a set x is usually represented by 'x or .cx
The complement of 'x or .cx
The complement of x is represented in the Venn-Euler diagram
U
x x
In set language, AxUxxAc ∉∈= ,:
The Algebra of Sets
The operations of union ∪ are loosely analogous to those of addition and multiplication
in number algebra. By this token we can apply the laws of algebra conveniently to sets
without loss of generality.
The Closure property
If X and Y are sets which are subsets of the universal set U then the following hold:
UYX ⊂∪ and .UYX ⊂∩
The analogy in number algebra; using those operations of + and x are R∈=+ 532 and
;632 Rx ∈= where R is the real number system. If the addition or multiplication of 2 and
3 gives some number that cannot be found in the real number system R , we say the
operation of + or x is not closed.
Similarly in set theory, the operations of union and intersection are closed.
The Commutative Law
UYYX ∪=∪ .XYYX ∩=∩ Parallel examples in arithmetic are 2332 +=+ and
.2332 XX =
Thus any two sets are commutative with respect to ∪ and .∩
The Associative Law
UZYXZYX )()( ∪=∪∪ and
ZYXZYX ∩∪=∩∩ )()(
Again, sets obey the associative law.
The Identity
In every day arithmetic, 10110 =+=+ and ,33113 == XX are two correct solutions. The
zero, in the first case is called the additive identity; while 1 in the second case is called
the multiplicative identity.
By a similar analogy , every set has quantities { }and U with the property that:
(i) { } { } XXX =∪=∪
(ii) XXUUX =∩=∩
Thus, { }is the identify with respect to union ∪ and U is the identity with respect to
intersection .∩
Inverse
In the set of real number ,R
0)()( =+−=−+ aaaa and .111 == XaaaXa This a number x operated on its inverse gives
identity. i.e., x Inverse = identity.
Similarly in set theory, every set has an inverse with respect to the operations of ∪ and ∩
(i) UUXXX ==∪ '' and
UXUUX =∪=∪
(iii) { }=∩=∩ XXXX '' and
{ } { } { }==∩ XXX '
The distributive Law
)()()( ZXYXZYX ∩∪∩=∪∩ and
)()()( ZXYXZYX ∪∩∪=∩∪
The operation of union is distributive over the operation of intersection and vice versa.
The Laws of complementation
(i) ∪=∪ 'XX and
(ii) XX =)'(
(iii) ')'( YXYX ∩=∪
(iv) '')' YXYX ∪=∩
(ii) – (iv) are called de Morgan’s Laws
Number set
Definition: Let { }10,9,8,6,4,3=A be a set, then we say set A has 6 elements, and We
write 6)( =An . If { }6,5,4,3,2=B , then 5)( =Bn and 8)( =∪ BAn .
Therefore, if A and B are disjoint,
)()()( BnAnBAn +=∪
If A and B are not disjoint )()()()( BAnBnAnBAn ∩−+=∪
This could be verified using the Venn diagram that
)()()(
)()()()()(
CBAnCBnCAn
BAnCnBnAnCBAn
∩∩+∩−∩−
∩−++=∪∪
APPLICATION OF SET THEORY The concept of the set theory could be applied to solve technological, industrial,
economic, social, educational and political problems.
Example 1: During the orientation programme of the newly admitted students of the
Federal University of Agriculture, Abeokuta, 18 students ate white-rice, 25 ate fried-
rice, 23 ate beans, 9 ate white and fried-rice, 10 ate fried-rice and beans and 6 ate
white-rice and beans. If there were 50 students altogether and 5 students did not take
any of the three meals. How many students ate
(i) all the meals
(ii) only fried-rice
(iii) beans but not fried-rice
(iv) white-rice and beans but not fried-rice
solution (1st method) :Let W,F, and B denote the students who ate white-rice, fried-
rice and beans respectively in the following Venn diagram. Let x denote those who
ate all the three meals and µ , the universal set.
Venn diagram
From the Venn diagram, we have xWBFn −=′∩∩ 10)(
xBFWn −=′∩∩ 9)(
xFBWn −=′∩∩ 6)(
Therefore,
x
xxxBFWn
+=
−−−−−=′∩′∩
3
])6()9(18[)(
Similarly,
x
xxxBWFn
+=
−−−−−=′∩′∩
6
])10()9(25[)(
and
x
xxxFWBn
+=
−−−−−=′∩′∩
7
])10()6(23[)(
Recall that the total number of students who did not eat any of the three meals is 5,
hence, the number of the students that ate one or two or the three meals =50-5=45.
Therefore,
4
4145
)7()10()6()6()9()3(45
=⇒
+=⇒
++−+++−++−++=
x
x
xxxxxxx
Hence,
(i) 4 students ate the three meals
(ii) 10 students ate only fried-rice
(iii) 13 students ate beans but not fried-rice
(iv) 2 students ate white-rice and beans but not fried-rice.
2nd
method Given that n(W)=18, n(F)=25, n(B)23
45550)(
6)(
10)(
9)(
=−=∪∪
=∩
=∩
=∩
BFWn
BWn
BFn
FWn
Using the relation
)()()(
)()()()()(
BFWnBFnBWn
FWnBnFnWnBFWn
∩∩+∩−∩−
∩−++=∪∪
Therefore,
4)(
)(106923251845
=∩∩⇒
∩∩+−−−++=
BFWn
BFWn
Which provides solution to (i), others can also be obtained using (i).
Example 2: A survey of the number of university students who read the three
newspapers was conducted. Their preferences for the new Nigerian, the Daily Times
and the Pioneer newspapers are given below:
48% read the New Nigerian
26% read the daily Times
36% read the Pioneer
8% read the New Nigerian and Pioneer
5% read the daily Times and Pioneer
4% read the New Nigerian and the Daily Times
4% read none of the newspapers
Calculate the percentage of those who read
(i) all three newspapers
(ii) only the Pioneer
(iii) only the New Nigerian
Solution: (i) Let N,D and P denote the percentage of the students who read the New
Nigerian, the Daily Times and the Pioneer newspapers respectively. Suppose 100
students were interviewed, then
n(N)=48, n(D)=26, n(P)=36
4)(
5)(
8)(
=∩
=∩
=∩
DNn
PDn
PNn
Since 4% of the students read neither N,D, nor P, then
964100)( =−=∪∪ PDNn
3)(
)(9396
)(45836264896
=∩∩⇒
∩∩+=⇒
∩∩+−−−++=
PDNn
PDNn
PDNn
Hence, 3 students read all the newspapers.
(iii) since 3)( =∩∩ PDNn and 5)( =∩ PDn
therefore
2623536)(
538)(
235)(
=−−−=⇒
=−=∩′∩
=−=∩∩′
Pn
PDNn
PDNn
Therefore, 26% read only Pioneer
(iii) those who read only the New Nigerian =48-5-3-1=37%
SURDS
Definition 1: Any number which can be expressed as a quotient n
m of two integers
( 0≠n ), is called a rational number. Any real number which is not rational is called
irrational. Irrational numbers which are in the form of roots are called surds. For
example, 2 , 3 , 5 ,π and 23 are irrational numbers while 16 , 83 and 325 can be
expressed in rational form.
Definition 2: A general surd is an irrational number of the form n ba , where a is a
rational number and n b is an irrational number , while n is called a radical.
RULES FOR MANIPULATING SURDS
(i) bcabcba )( +=+ . This is the addition law of surds with the same radicals.
(ii) dcadcda )( −=− . This is the subtraction law of surds with the same radicals.
(iii) baab .= .
(iv) bdacdcba =)).(( .
(v) b
a
b
a=
(vi) d
b
c
adcba =÷ )()(
(vii) 22)( aaa ==
(viii) nn aa =)(
(ix) m
m
aa
1=−
(x) m
ma
a=
−
1
Simplification of surds
Example; Simplify the following (i) 75 (ii) 80 (iii) 18 (iv) 60
Solution: Using rule 3
(i) 75 = 353.25325 ==×
(ii) 80 = 545.16516 ==×
(iii) 18 = 232.929 ==×
(iv) 60 = 15215.4154 ==×
Addition and subtraction of surds
Example; Simplify the following (i) 321850 +− (ii) 452080 −+ (iii)
6328 +
Solution: Using rule 1 and 2
(i) 321850 +−
21629225 ×+×−×=
242325 +−=
26=
(ii) 452080 −+
5954516 ×−×+×=
535254 −+=
53=
(iii) 6328 +
7974 ×+×=
7372 +=
75=
Rationalization of surds
A surd of the form 2
3 cannot be simplified, but
3
2 can be written in a more convenient
form. Then, we multiply the numerator and denominator of 3
2 by 3 . Such that
3
32
3
3
3
2
3
2=×= . This process is called rationalization.
Useful hints on rationalization of surds
(i) aaa =.
(ii) bababa −=−+ ))((
(iii) byaxbyaxbyax 22))(( −=−+
(iv) byxbyxbyx 22))(( −=−+
(v) the conjugate of ba + is ba −
Example; (i) Rationalizebn
a (ii) if 732.13 = , find the value of
3
2 correct to 3
significant figure (iii) Express 2332
638
+
− in the form 23 nm + where m and n are
rational numbers
(iv) Express 2)453(
2
− in the form cba + , where a and b are rational numbers
Solution:
(i) b
ba
bb
ab
bb
ba
bn
an n
n
n
n
n
n
n nn
n n 1
11
1
1
1
.
−
−
−
−
−
===
(ii) 16.13
732.12
3
32
3
3
3
2
3
2=
×==×= correct to 3 sig.figure.
(iii) 1812
129186224316
2332
2332
2332
638
3332
638
−
+−−=
−
−×
+
−=
+
−
73
17
2733
17
6
242334
6
318218224316
=−=⇒
+−=−
−=
−
+−−=
nandm
(iv)
5,841
48,
841
122,5
841
48
841
122
)52461)(52461(
)52461(2
52461
2
1652445
2
)453(
22
===⇒+
+−
+=
−=
+−=
−
cba
Equations involving surds
Example; (i) Solve the equation 1)4()13( =+−+ xx
(ii) simplify 625 + (iii) Evaluate 249 −
Solution:
(i) )4(1)13(1)4()13( ++=+≡=+−+ xxxx (1)
squaring both sides of (1), we have
)4(2
)4(24242513
4)4(2113])4(1[13 2
+=−⇒
+=−⇒+++=+⇒
++++=+⇒++=+
xx
xxxxx
xxxxx
(2)
squaring both sides of (2) again yields
solutiontheisx
xxxif
solutionnotxxxif
xorxxxxx
5
1)4()13(,5
)(1)4()13(,0
50054)2( 22
=⇒
=+−+⇒=
−=+−+⇒=
==⇒=−⇒+=−
(ii) Let yx +=+ 625 (1)
Squaring both sides of (1), we obtain
xyyx
xyyx
=+=⇒
++=+
6,5
2625
By inspection, 2,3 == yx
23625 +=+⇒
(iii) Let yx −==− 249 (1)
The conjugate of (1) is
yx +=+ 249 (2)
Squaring both sides of (1), we have:
xy2249 −=−
9=+⇒ yx (3)
Multiplying (1) and (2), we obtain
72.1699 =−=− yxx (4)
From (3) and (4)
9=+⇒ yx
7=− yx
162 =∴ x 1,8 == yx
EXERCISES
(1) (i) 405 (ii) 98 (iii) 1227 −= (iv) ( )257 − (v)
13
1
− (vi)
27
3
−
(vii) 27
3
− (vii)
13
13
+
− (viii)
122
322
−
+
(2) If ,32 +=a Find the value of a
a1
−
(3) Given 32
1
−a ,
32
1
+b , find 22
ba +
(4) Find the positive square roots of the following :
(i) 2619 + (ii) 71243 +
(5) If ( )512
1−=x , express xx 34 3 − in its simplest form.
INDICES
Definition 1: The product of a number with itself called the second power of the
number, while the number, while its triple product is called third power of the number
and its m factors product is called mth power of the number e.g. 2aaxa = , 3
aaxaxa = , m
axmaxax =...
Definition 2: The number which expresses the power is called the index or the exponent
of the power of a number e.g
The index of 22 =a
The index of 33 =a
The index of mam =
RULES OR LAWS OF INDICES
Given two positive integers nm, such that nm < .
(1) nmnmaxaa
+=
Since ( ) ( )xnaxaxxxmaxaxxaa nm ......=
[ ( ) ] nmanmxaxax +=+...
(2) )...( nmaxaxax −=
nma
−=
LOGARITHMIC EQUATIONS
Example 1: Solve the equations
(i) 42 93 += xx
( ) 422 33+=⇒
xx
( )422 33 += xx
( )422 +=⇒ xx
0822 =−=⇒ nx 0)2)(4( =+− xx
4−x or 2=x
Example 2: Solve the equations
(ii) 113 53 ++ = xx Taking the log; of both sides
( ) 1
10
13
10 5log2log ++⇒ xx
51213 1010 LogxLogx +=+⇒
( ) 2553 101010 LogLogxLogLog −=−⇒
( ) 25583 101010 LogLogxLogLog −=−⇒
( ) 2558 101010 LogLogxLogLog −=−⇒
2
5
8
2
5
58
2510
10
10
1010
1010 Log
Log
Log
LogLog
LogLogx −=
−
−=⇒
95.12041.0
3979.0==⇒ x
(3) mnnm aa =)(
xnxxaaa mmnm ...)( =
nxmaxaxxxmaxax )......()...( times.
mnaxmnaxax =...
(4) ( ) an
av =
1
( ) nxn
xan
an
a ...111
annn
a ...111
+++
Similarly, ann
a =1
( ) mna
n
ma =
( ( )mmmananan
n
ma ===
(5) 10 =a , If nm =
n
n
aa
1=−
Examples: Evaluate (i) 4
3
)81( (ii) 4
5
)16(
−
Solution
(i) 4
3
)81( 273)3(531441481( 34
3
43 =====
(ii) 4
5
)16(
−
32
1
2
1
4
5)2(
1
4
===
Exercises
(1) Show that ax
axax
+
−=−
(2) Evaluate (Godman).
LOGARITHMS
Definition: The logarithm of a tve no N to the base a is defined as the power of a
which is equal to N , such that if
Nax =
NLogx a=
Since 101 == aaa δ
1=⇒ aLog a and 01 =aLog
LAWS OF LOGARITHMS
(1) BLogALogAB aaa +=)(log
(2) BLogALogB
Aaaa −=log
(3) ABLogA a
B
a =)(log
Example: Evaluate:
(i) 93Log
(ii) 634Log
(iii) 2993 3
3 =⇒ Log
(iv) 364,644 4
3 =⇒= Log
Example: Use the table to evaluate:
(i) 524.24771.0
2041.1
3
1616
10
10
3 ===Log
LogLog
Since from the transformation rule NbLogLogNLog 6109 =
If y
b bNNLogy == , yy
a bybLogNLog ===⇒ )(
NbLogLogNLog baa =
If we put aN = in (*)
1==⇒ abLogLogaLog baa
bLogbLog
a
a
1=⇒ (**)
Another form of (*) is aLog
NLogNLog
b
b
a =
Example: Show that:
)1(2)(2
222
a
xLogxxLog aa −+=−
Solution:
[ ])1()(2
2222
a
xaLogxaLog aa −+=−
])1(2
22
a
xLogaLog aa −+=
])1(22
2
a
xLog a −+=
COURSE CODE: MTS 105
COURSE TITLE: Algebra and Trigonometry for Biological Sciences
NUMBER OF UNITS: 3 Units
COURSE DURATION: Three hours per week
COURSE DETAILS:
Course Coordinator: Drs I.A.Osinuga and I.O.Abiala
Email: [email protected] ; [email protected]
Office Location: Departmental Office
Other Lecturers: Prof.Oguntuase;Drs Olajuwon, Adeniran, Akinleye and
Mrs Akwu
COURSE CONTENT:
Matrix Algebra: Matrices. Algebra of Matrices. Determinant of a Matrix. Properties. Inverse
of a matrix. Solution of linear system of Equations.
COURSE REQUIREMENTS:
This is a compulsory course for all students in the University. In view of this, students are
expected to participate in all the course activities and have minimum of 75% attendance to be
able to write the final examination.
READING LIST:
1. Liadi, M.A. and Osinuga, I.A. A First Course in Mathematical Methods for Scientists and
Engineers. Ibadan: Rasmed Publications (Nig.) Limited, 2011.
2. Tranter,C.J. and Lambe, C.G., Advanced Level Mathematics(Pure and Applied). St.Paul:
The English Universities Press Limited, 1996.
3. Talbert, J.F. and Godman, A., Additional Mathematics for West Africa,
United King: Longman Group Limited, 1984.
4. Akinguola, R.O. et al., Introductory Mathematics I(for Social and Management
Sciences). Ago-Iwoye: CESAP Publication Unit,OOU. 1998.
MATRICES
Introduction
This chapter introduces an important branch of applied mathematics called matrices. The
studies of matrices are of immense use because they provide short cut methods of calculations
and this increases its suitability for practical problems.
We shall begin with the definition of matrices and related concept. We later discussed
types of matrices and some operations on matrices and their applications in the solution of
system of linear equations.
Definitions
Matrix is defined as a rectangular array of numbers or variables enclosed by a pair of
bracket whose position in the matrix is significant. The individual entries are referred to as the
elements of the matrix. The dimensions of a matrix (m x n) are defined as the numbers of rows
m and columns n which is read as “m by n”. If the number of rows equals the number of the
columns in a matrix, such a matrix is referred to as a square matrix. In a column vector
composed of a single column, such that the dimension is m x 1; if a matrix is a single row with
dimension 1xn, it is a row vector.
Illustration 1:
Given A = a11 a12 a13… a1n and B = 2 2 3
a21 a22 a23… a2n 1 2 4
… … … …
… … … …
… … … …
am1 am2 am3…amn
Here A is a general array with any number m of rows and any number n of columns called m x n
matrix.
Here B is a 2 x 3 matrix with b11 element as 2 and b13 as 3
Algebra of Matrices
Addition and Subtraction
The addition (subtraction) of matrices can only be carried out if the two matrices are of the same
dimension. The sum of two m x n matrices A = aik and B = bik is the m x n matric C = cik with
element cik = aik + bik written as C = A + B.
Illustration 2:
If A = 2 4 5 and B = 1 0 2
6 2 8 3 4 5
Then A + B = 2 4 5 + 1 0 2 = 3 4 7
6 2 8 3 4 5 9 6 13
Similarly,
A – B = 2 4 5 - 1 0 2 = 1 4 3
6 2 8 3 4 5 3 -2 3
Scalar Multiplication
Ordinary numbers such as 3, -2, 6, 4 etc. are called scalars. Multiplication of a matrix by a scalar
involves multiplying each element of the matrix by the scalar.
Illustration 3:
Given k = 3 and B = 1 2 1
2 2 4
kB = 3(1) 3(2) 3(1) = 3 6 3
3(2) 3(2) 3(4) 6 6 12
Matrix Multiplication
Two matrices A and B are conformable for multiplication if the number of columns in A
equals the number of rows in B. For instance, a 3 x 2 matrix can be multiplied with a 2 x 3
matrix since they are conformable (i.e. number of columns in the first equal number of rows in
the second matrix) but a 2 x 3 cannot be multiplied with another 2 x 3 matrix.
Illustration 4:
Let A = 2 3 and B = 1 2 8
1 2 2 1 0 calculate AB
1 5
Solution:
First test for the conformability of the two matrices. The number of columns in A equal the
number of rows in B, 2 = 2; the matrices are conformable for multiplication; and the dimensions
of the product matrix AB will be 3 x 3
AB = 2 3 1 2 8 = 2(1) + 3(2) 2(2) + 3(1) 2(8) + 3(0)
1 2 2 1 0 1(1) + 2(2) 1(2) + 2(1) 1(8) + 2(0)
1 5 1(1) + 5(2) 1(2) + 5(1) 1(8) + 5(0)
∴ AB = 8 7 16
4 4 8
11 7 8
Illustration 5:
If A is the 3 x 2 matrix in illustration 4 above, find the product AC where
C = 2 4
3 2
Solution:
The number of columns in A equals number of rows in C, 2 = 2; the matrices are
conformable for multiplication. Hence the results is as follows:
AC= 2 3 2 4 = 2(2) + 3(3) 2(4) + 3(2)
1 2 3 2 1(2) + 2(3) 1(4) + 2(2)
1 5 1(2) + 5(3) 1(4) + 5(2)
∴ AC = 13 14
8 8
17 4
Laws in Matrix Algebra
It is important to note that some elementary laws of algebra also apply to matrices since
their sum and difference is defined directly in terms of the addition and subtraction of their
elements.
Also, matrices satisfy the associative and distributive laws of multiplication
Commutative Laws
Addition and subtraction of matrix are both commutative because it merely involves the
summing of corresponding element of two matrices.
A + B = B + A
A + (-B) = (-B) + A written as A – B = -B + A
Clearly, AB ≠ BA (i.e. matrix multiplication is non commutative) since the order of matrices
representing two products are different. However, Scalar Multiplication is commutative (i.e. kA
= Ak)
Note: A – B ≠ B – A
Associative Laws
Matrix addition (subtraction) as stated earlier involves merely adding (subtracting) of
corresponding element of two matrices and it does not matter in which sequence the matrices are
added. For the same reason matrix addition is also associative.
(A + B) + C = A + (B + C)
The same is true of matrix subtraction if A – B is written as A + (-B). Moreover, if three
or more matrices are conformable for multiplication, the associative law will apply as long as
the matrices are multiplied in order of conformability. Thus,
(AB)C = A (BC)
Distributive Laws
Subject to the conditions of associative laws of multiplication, matrix multiplication is
also distributive.
A (B + C) = AB + AC
provided that the products are defined.
Illustration 6:
Given A = 2 3 B = 4 2 verify (i) A + B = B + A
5 2 1 0 (ii) A – B = -B + A
Solution
(i) A + B = 2 3 + 4 2 = 6 5
5 2 1 0 6 2
B + A = 4 2 + 2 3 = 6 5
1 0 5 2 6 2
∴ A + B = B + A i.e. commutative law of matrix addition holds
(ii) A – B = 2 3 - 4 2 = -2 1
5 2 1 0 4 2
-B + A = -4 -2 + 2 3 = -2 1
-1 0 5 2 4 2
∴ A – B = -B + A
Illustration 7:
Given A = 2 1 0 B = 1 2 C = 2 4 5
4 1 6 3 4 1 2 1
1 1
Find (i) (AB)C (ii) A(BC)
Solution:
(i)AB = 2 1 0 1 2 = 2(1) + 1(3) + 0(1) 2(2) + 1(4) + 0(1)
4 1 6 3 4 4(1) + 1(3) + 6(1) 4(2) + 1(4) + 6(1)
AB = 5 8
13 18
(AB) C = 5 8 2 4 5 = 18 36 33
13 18 1 2 1 44 88 83
(ii) BC = 1 2 2 4 5 = 4 8 7
3 4 1 2 1 10 20 19
1 1
A (BC) = 2 1 0 4 8 7 = 18 36 33
4 1 6 10 20 19 44 88 83
∴ Matrix Multiplication is associative
Types of Matrices
Null Matrices
Any (m x n) matrices with all its elements equal to zero is called a null matrix. As with
ordinary number, addition or subtraction of null matrix to (from) a matrix leaves the original
matrix unchanged and a null matrix is obtained if a matrix is multiplied by a null matrix.
Diagonal Matrices
A square (i.e. matrix with equal number of rows and columns) whose elements
everywhere are zero except at the leading diagonal is called a diagonal matrix. An example is
A = x 0 0
0 y 0
0 0 z
Upper and Lower Triangular Matrices
A square matrix in which all the elements below the leading diagonal are zero is called
the upper triangular matrix while the one in which all the elements above the leading diagonal
are zero is called the lower triangular matrix. They are usually denoted by U and L respectively
e.g.
2 3 1 2 0 0
U = 0 4 2 L = 3 4 0
0 0 1 1 2 1
Unit Matrices
Is a diagonal matrix with all its diagonal elements equal to unity and zero everywhere
else. It is usually denoted by letter I. For example:
1 0 0
I = 0 1 0 is a 3 x 3 unit matrix.
0 0 1
Remarks: Clearly, there is similarity between the unit matrix I and the number 1. As with
ordinary number where a number is multiplied by one equals itself so with unit matrices.
A matrix multiplied by the unit matrix equal itself provided the product is defined, i.e. AI
= IA = A.
For example,
1 0 1 = 1
0 1 3 3
Multiplication of the unit matrix by itself leaves the matrix unchanged:
I = I2 = I
3 = I
4 = …
Transposed Matrix
The resulting matrix obtained by interchanging rows and columns of a matrix ix called a
transposed matrix. Supposed A is given by
1 0 2
A = 4 2 3
2 3 4
Then the transpose of A is given by
AT = 1 4 2
0 2 3
2 3 4
Symmetric and Skew – Symmetric Matrices
A real square matrix A = ajk is said to be symmetric if it is equal to its transpose
AT = A i.e. ajk = akj (j, k = 1,…, n)
Whereas a real square matrix A = ajk is said to be skew – symmetric if
AT = -A i.e. ajk = -ajk (j, k = 1,…, n)
Determinant of a Matrix
The determinant of a matrix A is a number or scalar defined only for square matrices and
usually denoted by |A| or det (A). It is calculated from the products of the elements of the
matrix. If the determinant of a matrix equal zero, the matrix is termed singular; otherwise it is
said to be non-singular.
The determinant |A| of a 2 x 2 matrix, called a second-order determinant is defined as
follows:
Let A = a c , |A| = ad – cb
b d
Similarly, for a 3 x 3 matrix, the determinant called third-order determinant is calculated thus:
A = a d g
b e h
c f i
|A| = a e h + d (-1) b h +g b e
f i c i c f
Properties of a determinant
(1) The determinant of a square matrix is zero if every element of a row or column of such
matrix is zero.
(2) The determinant of a matrix equals the determinant of its transpose: |A| = |AT|
(3) If every elements of any row (or column) are multiplied by a constant, the value of the
determinant is multiplied by this constant: |kA| = k|A|.
(4) The value of a determinant is unaffected by interchanging the elements of all
corresponding rows and columns.
(5) The value of a determinant is zero if any two of its rows (or columns) are proportional.
(6) The value of a determinant is unchanged if any non-zero multiple of one row (or column)
are added (or subtracted) to the corresponding elements of any other rows (or columns).
Illustration 8:
Given A = 1 2 1
3 2 4
8 9 0
Find (i) Determinant of A
(ii) Determinant of the Transpose of A
and indicate which property of the determinant this illustrate.
Solution:
(i) |A| = 1(-36) – 2(-32) + 1(11) = -36 + 64 + 11 = 39
(ii) AT = 1 3 8 |AT| = 1(-36) – 3(-9) + 8(6) = 39
2 2 9
1 4 0
Since |A| = |AT|, this illustrates properties 2
Minor, Cofactors and Adjoint of a Matrix
Given a determinant of order 3 written as a square array of 9 quantities enclosed between
vertical bars.
D = a11 a12 a13
a21 a22 a23
a31 a32 a33
By deleting the ith row and kth column from the determinant D, we obtain a sub-determinant of
the matrix called a minor. Thus a minor |mik| is the determinant of the sub-matrix obtained by
deleting the ith row and kth column of the matrix.
Using a (3 x 3) matrix A = a11 a12 a13
a21 a22 a23
a31 a32 a33
Deleting the first row and first column in A, i.e.
a11 a12 a13
a21 a22 a23 we obtain |M11| = a22 a23
a31 a32 a33 a32 a33
Similarly deleting first row and second column
a11 a12 a13
a21 a22 a23 we obtain |M12| = a21 a22
a31 a32 a33 a31 a32
In the same manner we have
|M13| = a21 a22
a31 a32
A co-factor |Cik| is a minor multiplied by (-1)i+k
: thus
|Cik| = (-1)i+k
|Mik|
i.e. co-factor is a matrix in which every entries aik is replaced with its cofactor |Cik|. An adjoint
matrix is the transpose of a cofactor matrix.
Thus ifA = |C11| |C12| |C13|
|C21| |C22| |C23|
|C31| |C32| |C33|
Adjoint A = CT = |C11| |C21| |C31|
|C12| |C22| |C32|
|C13| |C23| |C33|
Illustration 9:
Given A = 3 -4 -6 find (i) The cofactor of matrix A
1 2 -4 (ii) Adjoint of A
2 -5 2
2 -4 - 1 -4 1 2
-5 2 2 2 2 -5
- -4 -6 3 -6 - 3 -4
-5 2 2 2 2 -5
-4 -6 - 3 -6 3 -4
2 -4 1 -4 1 2
-16 -10 -9
Cof.A = 38 18 7
28 6 10
(ii) Adjoint A = (Cof. A)T = -16 38 28
-10 18 6
-9 7 10
Inverse of a Matrix
In the matrix algebra, the inverse of a matrix A is denoted by A-1
and calculated only for
square, non-singular matrix such that
A-1
A = I = AA-1
It can easily be verified that multiplication of matrices and their inverses is commutative
and that multiplying a matrix by its inverse yields an identity matrix. Thus, the inverse of a
matrix has the property as the reciprocal in ordinary algebra.
In this section, we present a method of calculating the inverse of a matrix:
Adjoint method: The formula for finding the inverse of a given matrix A is given as:
A-1
= adjoint A / det. A
where det. A is the determinant of A.
Illustration 10:
Find the inverse of a matrix A = 2 3
5 -2
Solution:
|A| = -4 – 15 = -19
Cof. A = -2 -5
-3 2 then
(Cof. A)T = -2 -3
-5 2
A-1
= (Cof. A)T = -1/19 -2 -3
|A| -5 2
∴ A-1
= 2/19 3/19
5/19 -2/19
6.9 Solution of Systems of Linear Equations Using Matrix Inverse
Consider the following system of linear equation in unknowns x1, x2, …, xn:
a11x1 + a12x2 + . . . + a1nxn = b1
a21x2 + a22x2 + . . . + a2nxn = b2
. . .
. . .
. . .
an1x1 + an2x2 + . . . +annxn = bn
where the coefficient aix and bi (i, x = 1, 2, 3, … …, n) are known constants. This equation can
be written in the matrix form as:
a11 a12 … a1n x1 b1
a21 a22 … a2n x2 b2
… … … … = … or AX = B
… … … … …
… … … … …
an1 an2 … ann xn bn
Assuming B ≠ 0 and A to be non-singular matrix (i.e. A-1
exists).
Multiplying both sides of the equation by A-1
gives
A-1
AX = A-1
B (A-1
A = I)
∴ X = A-1
B
The solution of the equation AX = B is given by the equation X = A-1
B i.e. the product of the
inverse of the coefficient matrix A-1
and the column vector of constants B.
Illustration 11:
Solve the equation 4x + 5y = 0
2x + 3y = 0
Solution: First write the equation in matrix form
AX = B
4 5 x = 2
2 3 y 0
In order to find the inverse of of the coefficient matrix, proceed as follows:
A = 4 5 |A| = 2
2 3 ,
Cof. A = 3 -2 (Cof. A)T = 3 -5
-5 4 -2 4
A-1
= Adjoint A = (Cof. A)T = ½ 3 -5
|A| |A| -2 4
Thus X = A-1
B = 3/2 -5/2 2 = 3
-1 2 0 -2
So the solution of the equation is x = 3, y = -2.
Cramer’s Rule
Cramer’s rule is rule that provides a shorthand method of solving linear equation. The
concept follows that if D ≠ 0, the system of linear equation has the unique solution given by:
Xi = Di
D
Where D1 is the determinant of a special matrix formed from the original matrix by replacing the
column of coefficient of xi with the column vector of constants, D is the determinant of the
coefficient matrix and xi is the ith unknown variable in a series of equation.
Illustration 12:
Solve by Cramer’s rule
2x1 + 3x2 = 8
5x1 – 2x2 = 1
Solution: Express the equations in matrix form:
2 3 x1 = 8
5 -2 x2 1
The determinant of matrix A is
D = 2 3 = -4 – 15 = -19
5 -2
To solve for x1 = D1/ D, replace a column 1, the coefficient of x1, with vector of constants B,
forming
D1 = 8 3 = -16 – 3 = -19
1 -2
And x1 = D1 = -19 = 1
D -19
Similarly, to solve for x2 = D2 / D, replace column 2, the coefficient of x2, with vector of
constants B, forming:
D2 = 2 8 = 2 – 40 = -38
5 1
Which gives x2 = D2 = -38 = 2
D 19
∴ x1 = 1 and x2 = 2
Remarks: Given systems of the three linear equations of the matrix form
a11 a12 a13 x1 b1
a21 a22 a23 x2 = b2
a31 a32 a33 x3 b3
By Cramer’s rule, the system of three linear equation yields the solution
x1 = D1 x2 = D2 and x3 = D3
D D D
Where D1 = b1 a12 a13 D2 = a11 b1 a13
b2 a22 a23 a21 b2 a23
b3 a32 a33 a31 b3 a33
and D3 = a11 a12 b1
a21 a22 b2
a31 a32 b3
Illustration 13:
Solve by Cramer’s rule
-x + 3y – 2z = 7
3x + 3y = -3
2x + y + 2z = -1
Solution: Rewrite the equations as:
-1 3 -2 x = 7
3 0 3 y -3
2 1 2 z -1
The find the determinant D,
-1 3 -2
D = 3 0 3 = -3
2 1 2
Also,
7 3 -2 -1 7 -2
D1 = -3 0 3 = -6, D2 = 3 -3 3 = -3 and
-1 1 2 2 -1 2
-1 3 7
D3 = 3 0 3 = 9
2 1 -1
Solving for x1, x2 and x3; we have
x1 = D1/D = -6/-3 = 2; x2 = D2/D = -3/-3 = 1 and x3 = D3/D = 9/-3 = -3
∴ x1 = 2, x2 = 1 and x3 = -3
Exercises:
COURSE CODE: MTS 101
COURSE TITLE: Algebra
NUMBER OF UNITS: 3 Units
COURSE DURATION: Three hours per week
COURSE DETAILS:
Course Coordinator: Dr. I.A.Osinuga and Miss A.D.Akinola
Email: [email protected] ; [email protected]
Office Location: Departmental Office
Other Lecturers: Prof.Oguntuase;DrsAdeniran,Akinleye,Omeike;Mr.Ilodje
and Mrs Akwu
COURSE CONTENT:
Sequences and Series: AP and GP, Means, nth term and limits. Binomial Theorem for any
index. B inomial series.
COURSE REQUIREMENTS:
This is a compulsory course for all students in the University. In view of this, students are
expected to participate in all the course activities and have minimum of 75% attendance to be
able to write the final examination.
READING LIST:
5. Tranter,C.J. and Lambe, C.G., Advanced Level Mathematics(Pure and Applied). St.Paul:
The English Universities Press Limited, 1996.
6. Talbert, J.F. and Godman, A., Additional Mathematics for West Africa,
7. Akinguola, R.O. et al., Introductory Mathematics I(for Social and Management
Sciences). Ago-Iwoye: CESAP Publication Unit,OOU. 1998.
SEQUENCES AND SERIES
Sequences
Sequences can be defined as a succession of things or number occurring in some
particular order that can be determined by a definite law. For example, the sets (i) 1, 3, 5,
7, 9, …
(ii) 2, 4, 6, 8, 10, …
(iii) 1, 4, 9, 16, 25, …
(iv) 4, -1, -6, -11, …
are all sequences and each number is a term of the sequence. The dots … indicate that the
sequence continues indefinitely.
Series
Series on the other hand can be described as the resulting expression when the terms of a
sequence are linked together with signs of subtraction or addition.
Examples of series include:
(i) 1 + 3 + 5 + 7 + 9 + …
(ii) 1 + 4 + 9 + 16 + 25+ …
(iii) 4 – 1 – 6 – 1 – ….
Observe that when the terms of a finite sequence are added together, we obtain a finite series.
Thus, the series (ii) contains a finite sequence of 5 terms to give a finite series but series (i) and
(iii) continues infinitely (i.e. infinite series).
The nth Term
It is possible to give a formula for the nth term of a sequence. More so, it is easy to write
down any term in the sequence once the nth term or general term is known. The general term are
usually denoted by Tn or Un. We shall illustrate with examples how to write down the nth term
formula of some of the above sequences.
Illustration 1:
Write down the nth term formula for the sequences (i) and (iv) above.
Solution: The first sequence is 1, 3, 5, 7, 9 …
We construct a table in order to see the pattern of the sequence
Table 1
Number 1st 2nd 3rd 4th 5th …. 10th …. nth
of term
Term 1 1+2 1+2+2 1+2+2+2 1+2+2+2+2 … …
1+2(1) 1+2(2) 1 + 2(3) 1 + 2(4) 1 + 2(9) 1+2(n-1)
Hence, the nth term of the sequence is given by:
1 + 2(n – 1) = 1 + 2n – 2 = 2n – 1.
The sequence (iv) is 4, -1, -6, -11, …
Table 2
No of term 1st 2nd 3rd 4th … 10th … nth
Term 4 4-5 4-5-5 4-5-5-5 … …
4-5(1) 4-5(2) 4-5(3) 4-5(9) 4-5(n-1
∴ The nth term is 4 – 5 (n -1) = 4 – 5n + 5 = 9 – 5n
Illustration 2:
Write out the first 4 terms of the sequence whose nth term is (i) n2 (ii) 2
n-1
Solution:
(i) Given nth term to be n2; Then
1st term becomes 1
2 = 1
2nd
term becomes 22 = 4
3rd
term becomes 32 = 9
4th
term becomes 42 = 16
∴ The first 4 terms are 1, 4, 9, 16
Note that the nth term formula n2 is general term of the sequence (iii) given above
(ii) Given nth term to be 2n-1
. Then
1st term gives 2
1 – 1 = 2
0 = 1
2nd
term gives 22 – 1
= 21 = 2
3rd
term gives 23 – 1
= 22 = 4
4th
term gives 24 – 1
= 23 = 8
∴ The first four terms are 1, 2, 4, 8
Progressions
If a sequence of terms is governed by a definite law, the terms are said to form a
progressions. Among the various forms of progressions, we have includes Arithmetic,
Geometric and Harmonic progressions.
We shall however lay emphasis on arithmetic and geometric progression
Arithmetic Progression (AP)
In an Arithmetic Progression, the sequences of terms differ from one another by a
constant quantity which may be positive or negative. This constant quantity is called the
common difference (d). For example, the sets
(i) 2, 4, 6 8, 10, …
(ii) 5, 9, 13, 17, 21, …
(iii) 4, -1, -6, -11, …
are all arithmetic progressions or arithmetic sequences. Thus, the common differences (d) in the
three sequences given above are 2, 4, and -5 respectively.
Generally, if we denote the first term in AP by ‘a’ the nth term of an AP is denoted by Tn
given by:
Tn = a + (n – 1) d
Illustration 3: Given the sequence 2, 5, 8, 11, …
Find (i) the formula for the general term
(ii) the 28th
term of the sequence
Solution:
(i) a = 2, d = 3
Tn = a + (n – 1) d = 2 + (n – 1) 3 = 3n – 1
(ii) T28 = a + (28 – 1)d = a + 27d = 2 + 27(3) = 83
Illustration 4:
The 5th term of an AP is -10 and the 10th
term is -25. What is the sequence?
Solution: T5 = a + 4d = -10 … (1)
T10 = a + 9d = -25 … (2)
Subtracting (2) from (1)
-5d = 15 i.e. d = -3
Then from (1) a – 12 = -10 i.e. a = 2
Hence the AP is 2, -1,-4, -7, …
Illustration 5:
How many terms has the AP, if the first and the last terms of an AP with d = 2 are 0 and 10
respectively?
Solution:
a = 0, Tn = 10, d = 2
Tn = a + (n – 1)d
10 = 0 + (n – 1) 2 i.e. 10 = 2n – 2
or n = 6
∴ There are 6 terms of the sequences.
The Arithmetic Mean
In an AP, if the three consecutive terms are given as x, y, z. Then the common difference
d = y – x = z – y
i.e. y + y = z + x
or y = z + x
2
Therefore, y is called the arithmetic mean of x and z. The arithmetic mean of 5 and 17 is given
by 5 + 17 / 2 = 11
Illustration 6: Insert 2 arithmetic mean between 2 and 11
Solution: To do this, find 2 numbers x, y between 2 and 11 such that 2, x, y, 11 forms an AP
a = 2, a + 3d = 11, d = 3
The required numbers are 2, 5, 8, 11
The Sum of an AP
The nth term of an AP whose first term is a and common difference d has been given as Tn = a
+ (n – 1)d
Then T2 = a + d
T3 = a + 2d
T4 = a + 3d etc.
If the sum of the 1st n terms of an AP is denoted by Sn, we have
Sn = a + a + d + a + 2d + a + 3d + … + a + (n – 1) d … (1)
Write the series backward and it becomes
Sn = a + (n – 1) d + … + a + 3d + a + 2d + a + d + a … (2)
Adding each term of the series in (1) to the corresponding term of series (2) we have
2Sn = n [2a + (n – 1) d]
or Sn = n/2 [2a + (n – 1) d]
Illustration 7:
Find the sum of the first 20 terms of AP: 0 + 2 + 4 + 6 + 8 + 10 + …
Solution: Sn = n/2 [2a + (n – 1) d]
= 20
/2 [0 + (20 – 1) 2] = 10 (38) = 380
Exercise 1:
1. The given expression is the nth of a sequence. Write down the first six terms of each
sequence. (i) n (b) n + 1 (iii) 2n-2
(iv) 5 – 2n (v) (-1)n
2
(2) Find the possible expression for Tn for each of the following sequences and write down
the next three terms
(i) 1, ½, ⅓, ¼, 1/5, …
(ii) 2, 4, 6, 8, 10, …
(iii) 3, 9, 27, …
(3) Two sequences are defined by Tn = n2 – n + 4 and Vn = 2
n-1. Find for each sequence
which term is equal to 16.
(4) In a sequence given by Tn = a + bn, the 7th
and 15th
terms are 19 and 42 respectively.
Find the values of a and b.
(5) Tn is given for each sequence (i) 3n+1 (ii) n(n + 2) (iii) 2n3. Find T3 and T10
(6) Find the 23rd term of the AP: 3, -1, -5, -9, …
(7) If (i) a = 3, T12 = -41 find d
(ii) a = 2, d = 3, Tn = 59 find n
(iii) T10 = 18, d = 2, find a
(8) The 5th
and 10th
term of an AP are 8 and 18 respectively. Find the AP and its 15th
term.
(9) If 1, x, y, 19 are in AP; find x and y
(10) Insert 2 arithmetic mean between 3 and 18
(11) Find the arithmetic mean of
(i) -7 and 3
(ii) 3 and 13
(iii) 12 and 4
(12) Find the sum of 12 terms of the series 7 + 13 + 19 + …
(13) How many terms of the AP 0 + 2 + 4 + 6 + … make a total of 380?
(14) What is the nth terms of the AP: 10, 6, 2, -2, …
(15) How many terms of the AP: 3, 7, 11, … must be added together to produce a total of
300?
Geometric Progressions (GP)
In a geometric progression, each term in a sequence of terms is a constant multiple of the
preceding one. This constant multiple is referred to as the common ratio (r). Thus, for example
(i) 2, 4, 8, 16, …
(ii) 2, 1, ½, ¼, …
(iii) -3, 9, -27, 81, …
are geometric progressions with common ratio 2, ½ and -3 respectively.
If a is the first term of a GP and r is the common ratio, the sequence will be
T1 = a
T2 = ar
T3 = ar2
T4 = ar3
.
.
.
T10 = ar9
.
.
.
Tn = arn-1
Following this pattern, we observe that the nth term of a GP is
Tn = arn-1
Illustration 8:
Given the sequence -2, -4, -8, -16, …
Find (i) the 9th term of the above sequence
(ii) a formula for the nth term.
Solution:
(i) Hence a = -2 and r = 4/-2 = -8/4 = -2
T9 = ar8 = (-2) (-2)
8 = -512
(ii) Tn = (-2) (-2)n-1
= (-2)n
Illustration 9:
Find the nth term of the series 4 + 6 + 9 + 27/2 + …
Solution: In the above series a = 4, r = 3/2
T1 = a = 4
T2 = ar = 4 (3/2) = 6
T3 = ar2 = 4 (
3/2)
2 = 0
.
.
.
Tn = arn-1
= 4 (3/2)
n-1
∴ The nth term of the series is 4 (3/2)
n-1
Illustration 10:
The second term of a GP is 12 and the 9th term is 108. Find the two possible value of a (first
term) and r.
Solution: T2 = ar = 12 and T4 = ar3 = 108
Then T4 = ar3 = 108 = 9 or r
2 = 0 i.e = ±3
T2 ar 12
From the first equation, when r = 3, ar = 12 or 3a = 12 i.e. a = 4
Also, when r = -3, we have ar = 12 or -3a = 12 i.e. a = -4
∴ The GP is either 4, 12, 36, 108, … or -4, 12, -36, 108, …
The Geometric Mean
The Geometric Mean follows the same procedure as in arithmetic mean. Given three
consecutive terms x, y, z; the common ratio is
y = z or y2 = xz i.e. y = √xz
x y
which is the condition for the three consecutive terms x, y, z to form a GP. The geometric mean
of two number is the square root of their products e.g. the geometric mean of 3 and 27 is √3 x 27
= √81 = 9
Illustration 11: Insert 2 GM between 1 and 125
Solution: We find numbers x, y such that 1, x, y, 125 forms a GP. The terms are a, ar, ar2 and
ar3.
a = 1, ar3 = 125 or r = 5
Hence x = ar = 5, y = ar2 = 25
∴ The required Geometric Mean are 5, 25
The Sum of a GP
Generally, let Sn be the sum of n terms of the GP with first term a and common ratio r.
Since the nth term is arn-1
, so that
Sn = a + ar + ar2 + … + ar
n-1 … (1)
Multiply (1) by r, we have
rSn = ar + ar2 + ar
3 + … + ar
n … (2)
Subtracting (2) from (1) becomes
Sn – rSn = a – arn
Sn (1 – r) = a (1 – rn) or Sn (r – 1) = a (r
n – 1)
Therefore,
Sn = a (rn – 1) = a (1 – r
n), r ≠ 1
r – 1 1 – r
The first form is more suitable if r > 1, the second form when r <1. If r = 1, the formula cannot
be used. In this case however, the sum of the series is na.
Illustration 12: Find the sum of the first 6 terms of the GP 1, 3, 9, 27, …
Solution: In the GP, a = 1, r = 3
S6 = a (rn – 1), r > 1
r – 1
S6 = 1 [(3)6 – 1] = 3
6 – 1= 728= 364
3 – 1 2 2
Remarks:
Observe that when r is very small (i.e. approaching zero) and n is sufficiently large the sum of
GP to infinity of a GP is
S = a___
1 – r
Since the limit of rn → 0 and n →∞
Exercise 2:
Write down the term indicated in each of the following Geometric Progression
1. 6, 3/2, 3/8, …, 7th
2. 4, 12, 36, … , 15th
3. -3, 9, -27, …, 6th
4. 5, -5/2, 5/4, … , 5th
5. 9, -3, 1, … , 10th
What is the nth term of the following GP
6. 2, 4, 8, …
7. 6, 18, 54, …
8. y, 2yz, 4yz2, …
9. 1000, 100, 10, …
10. 28, -14, 7, …
11. The 2nd and 6th terms of a GP (r > 0) are 4 and 64 respectively. Find the sum of first 5
terms.
12. If 1, x, y, 27 are in GP, find x and y
13. Find the geometric mean between x and 4xy2
14. How many terms of GP 4 + 9 + … must be taken to make the sum exceed 5000?
15. Find the sum to infinity of the series 75 + 45 + 27 + …
Applications
Compound Interest
When money is saved in a bank, the interest is paid on this money saved with compound
interest. The interest is added to the principal at the end of each interval so that the successive
terms of the series forms a geometric progression.
In this case, the principal and the interest for each interval increases.
If the sum of money, P is invested for M years at K% per annum compound interest, the
amount S, after M years is given by the formula
S = P (1 + K)M
Illustration 13:
A trader saved N1,000 in a bank which pay interest at 4% per annum. Find the amount in the
trader’s savings account after a year, 2 years, 3 years, …, 6 years.
Solution: The Compounding Formula is S = P (1 + K)M
After 1 year S = 1000 (1 + 0.04)1 = N1040.00k
After 2 years S = 1000 (1 + 0.04)2 = N1081.60k
After 3 years S = 1000 (1 + 0.04)3 = N1124.86k
After 6 years S = 1000 (1 + 0.04)6 = N1265.32k
Using Tabular Form:
P for 1st year = 1000;
I for 1st year = 40;
Principal year = 1040 amount after 1st year
Interest for 2nd
year 41.6
Principal for 3rd
year 1081.60k amount after 2nd
year
Interest for 3rd
year 43.264
Principal for 4th
year 1124.864 amount after 3rd
year
Interest for 4th
year 44.99
Principal for 5th
year 1169.854 amount after 4th
year
Interest for 5th
year 46.794
Principal for 6th
year 1216.65 amount after 5th
year
Interest for 6th
year 48.66
1265.32 amount after 6th
year
When the present value is required given the future value the compounding formula, is
restated in terms of discounting to a present value as
P = S____
(I + K)M
Illustration 14:
How much will have to be invested now to produce N862 after 6 years with a 9½ compound
interest rate?
P = 862 ___
(1 + 19
/200)6 = 500
Exercise 3:
1. Find the amount that N5,000 becomes if saved for 5 years at 6% per annum compound
interest
2. A company invests N10,000 each year for 5 years at 9% compound interest. What will be
the fund be after 5 years?
3. How much will have to be invested now to produce N20,000 after 10 years with a 10%
Compound Interest Rate?
4. A firm rents his premises and the rented agreement provided for a regular annual increase
of N2650. If the rent in the first year is N8500. What is the rent in the tenth year?
5. A man borrows N9000 to buy a house at 8% per annum compound interest. He repays
N920 at the end of that year. How much does he still owe at the end of 3 years?
6. Find the sum of the first 20 terms of the AP 5, 11, 17, 23, …
7. 16 cows were slaughtered for sale in a market town during the first week of January 197
and four more cows during each subsequent week. Assuming thee are 4 weeks in each
month, determine the number of cows slaughtered for sale at the end of December of the
same year?
8. How much will N10,000 amount to at 8% per annum compound interest over 15 years?
9. (i) How much will N20,000 amount to at 3% per annum compound interest over 10 years
(ii) What compound rate of interest will be required to produce N5,000 after five years
with an initial investment of N4,000.
10. The profit of a supermarket shows an annual increase of 6%. Assuming the current
market trends continue, what will be the supermarkets annual profit in the 4th
year if the
first profit was N20,000? Find also the total profit for the first 4 years.
BINOMIAL SERIES
Permutation: A permutation is an arrangement of a number of objects in a particular order.
Combination: When a selection of objects is made with no regard being paid to order, it is
referred to as combination e.g. ABC, BAC, CAB are different permutation but the same
combination of letters.
Factorial Notation
n! = n (n – 1) (n – 2) (n – 3) . . . 3 . 2. 1
e.g. 6! = 6. 5. 4. 3. 2. 1
4! = 4. 3. 2. 1
Notations:
1. Permutation: nPr = n!__
(n – r)!
2. Combination: nCr = n!____
(n – r)! r!
Remarks:
(a) nC1 =
nCn-1
(b) n+1
Cr+1 = nCr +
nCr+1
(c) nPr = r!
nCr (The proof of the above is left as exercise for the reader)
Examples:
5P3 = 5! = 5! = 5.4.3.2.1 = 60
(5-3)! 2! 2.1
5C3 = 5! = 5! = 5.4.3.2.1 = 10
(5 – 3)! 3! 2!3! 2.1.3.2.1
2.2 BINOMIAL THEOREM
Observe that (a + b)2 = a
2 + 2ab + b
2. It might be difficult for some of us to expand (a +
b)3 and other higher powers of (a + b) without doing some serious work on paper; through long
multiplication.
But since a0 = 1
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a
2 + 2ab + b
2
(a + b)3 = a
3 + 3a
2b + 3ab
2 + b
3
(a + b)4 = a
4 + 4a
3b + 6a
2b
2 + 4ab
3 + b
4 etc.
If the coefficients are written alone, we have
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1 etc.
On studying the order closely, we observe that in each case, the first and last coefficients are
unity.
Furthermore, we noticed that each of the other coefficient (a + b)n+1
is the sum of the
corresponding coefficient, and their preceding one in the expansion of (a + b)n. Thus, we can lay
out coefficients for the successive powers in the form of a triangle called Pascal’s Triangle.
However, for large values of n, it might not be easy to use Pascal’s Triangle, thus to write
down the expansion of (a + b)n for large n, we use a formular for the coefficient of a
n-rb
r in the
expansion of (a + b)n. We shall denote this by the symbol n which alternatively leads to
Binomial Theorem.
(a + b)n =
nC0 a
nb
0 +
nC1 a
n-1 b
1 +
nC2 a
n-2 b
2 + … +
nCn a
0 b
n
Illustration 1:
(1) Expand (x + y)6 in descending power of n
Solution: There will be seven terms involving
x6, x
5y, x
4y
2, x
3y
3, x
2y
4, xy
5, y
6
Their coefficients obtained from Pascal’s triangle are respectively,
1 16 15 20 15 6 1
Thus, the required expansion is
x6 + 6x
5y + 15x
4y
2 + 20x
3y
3 + 15x
2y
4 + 6xy
5 + 1y
6
Similarly, we could verify that
(x + y)6 =
6C0x
6y
0 +
6C1x
5y +
6C2x
4y
2 +
6C3x
3y
3 +
6C4x
2y
4 +
6C5xy
5 +
6C6x
0y
6
= x6 + 6x
5y + 15x
4y
2 + 20x
3y
3 + 15x
2y
4 + 6xy
5 + y
6
Recall that:
nC0 = 1
nC1 = n,
nC2 = n (n – 1),
nC3 = n (n – 1) (n – 2)
2! 3!
However, we can also see that
(1 + x)n = 1 +
nC1x +
nC2x
2 +
nC3x
3 + … +
nCrx
r + … +
nCnx
n =
nCrx
r
Illustration 2: Expand (i) (1 – 2x)
4 (ii) (1 + x + x
2)
3 in powers of x
(i) (1 – 2x)4 = (1 + (-2x))
4
= 1 + 4C1 (-2x)
1 +
4C2 (-2x)
2 +
4C3 (-2x)
3 +
4C4 (-2x)
4
= 1 + 4 (-2x) + 6(4x2) + 4(-8x
3) + 16x
4
= 1 – 8x + 24x2 – 32x
3 + 16x
4
(ii) (1 + x + x2)
3 = [1 + (x + x
2)]
3
= 1 + 3(x + x2) + 3(x + x
2)
2 + (x + x
2)
3
= 1 + 3x + 3x2 + 3(x
2 + 2x
3 + x
4) + (x
3 + 3x
4 + 3x
5 + x
6)
= 1 + 3x + 6x2 + 7x
3 + 6x
4 + 3x
5 + x
6
Illustration 3: Find the coefficient of x8 in (x
2 +
2y/x)
10
Solution: The (r + 1)th term in the expansion of (x2 +
2y/x)
10 is
10
Cr (x2)
10-r (
2y/x)
r
10
Cr 2r y
r x
20-2r-r
Thus, for the term in x8, we have
10
Cr 2r y
r x
20-3r, 20 – 3r = 8, 12 = 3r ⇒ r = 4
∴The required coefficients is
10C4 2
4 y
4 = 10.9.8.7! 16y
4 = 210.16y
4 = 3360y
4
4! 6!
∴ The coefficient of x8 in (x
2 +
2y/x)
10 = 3360y
4
n
∑ r=0
BINOMIAL EXPANSION AND APPLICATIONS
Illustration 4: Obtain the first five terms in the expansion of (1 + x)½.
Hence, evaluate
√1.03 to 5 significant figures.
Solution: For this case n = ½ and so by using the binomial expansion
(1 + x)n = 1 + nx + n (n – 1) x
2 + n (n – 1) (n – 2) x
3 + …
1.2 1.2.3
(1 + x)½
= 1 + ½x + ½ (½ - 1) x2 + ½ (½ - 1) (½ - 2) x
3 + ½ (½ - 1) (½ - 3) x
4 + …
2 2.3 1.2.3.4
= 1 + ½x + ⅛x2 +
1/16x
3 –
5/128x
4 + …
To evaluate √1.03, we have
√1.03 = (1.03)½ = (1+0.03)
½ = 1 + ½(0.03) – ⅛ (0.03)
2 +
1/16(0.03)
3 –
5/128(0.03)
4
= 1 + 0.015 – 0.0001125 + 0.000001687 – 0.000000031
= 1.0148892
≈ 1.0149 to 5 significant figures.
Illustration 5: Find the values of a, if the coefficient of x2 in the expansion of
(1 + ax)4 (2 - x)
3 is 6.
Solution: (1 + ax)4 = 1 + 4 (ax) + 6 (ax)
2 + 4 (ax)
3 + (ax)
4
(2 – x)3 = [2 + (-x)]
3 = 2
3 (-x)
0 + 3.2
2 (-x)
1 + 3 (2) (-x)
2 + (-x)
3
= 8 – 12x + 6x2 – x
3
(8 – 12x + 6x2 – x
3) (1 + 4ax + 6a
2x
2 + 4a
3x
3 + a
4x
4) = (1 + ax)
4 (2 – x)
3
The coefficient of x2 is:
48a2x
2 – 48ax
2 + 6x
2 ⇒ 6 – 48a + 48a
2 = 6
8a2 – 8a = 0
8a (a – 1) = 0
⇒ a = 0 or a = 1
Illustration 6: Find the term independent of x in the expansion of (x – ½x)12
Solution:
The general term will be 12
Cr x12-r
(-½x)r =
12Cr x
12-r (
1/xr) (-½)
r
Now x12-r
(1/xr) = x
12-2r
If the term is to be independent of x, then 12 – 2 r = 0 or r = 6 and this term (the 7th term)
is then
12
C6 x6 (-½ x)
6 =
12C6 x
6 (
1/x6) (-½)
6
= 12! x 1 = 12.11.10.9.8.7 x 1 = 231
6!6! 64 1.2.3.4.5.6 64 16
∴ The term independent of x (or the constant term) is 231
/16.
Illustration 7: If x is so small that its fourth and higher powers may be neglected, show
that (1 + x)¼
+ (1 – x)¼ = a – bx
2, and find the numbers a and b.
Hence, by putting x = 1/16, show that (17
¼ + 15
¼) ≈ 3.9985
Solution:
(1 + x)¼ = 1 + ¼ x –
3/32 x
2 +
7/128 x
3 …
(1 – x)¼ = 1 – ¼ x –
3/32 x
2 –
7/128 x
3 …
(1 + x)¼ + (1 – x)
¼ = 1 + ¼ x –
3/32 x
2 +
7/128 x
3 … + 1 – ¼ x –
3/32 x
2 –
7/128 x
3 …
= 2 – 3/16x
2 = a – bx
2
⇒ a = 2, b = 3/16
Putting x = 1/16
(1 + x)¼ + (1 – x)
¼ = (1 +
1/16)
¼ + (1 –
1/16)
¼
= 17 ¼ + 15
¼ = 17
¼ + 15
¼ = 17
¼ + 15
¼
16 16 16¼ 16
¼ 2 2
= ½ (17¼ + 15
¼)
(1 + x)¼ + (1 – x)
¼ = ½ (17
¼ + 15
¼) = 2 –
3/16x
2
If x = 1/16
= ½ (17¼ + 15
¼) = 2 –
3/16 (
1/16)
2
= ½ (17¼ + 15
¼) = 8189
4096
17¼ + 15
¼ = 8189 x 2 = 3.998535156
4096
∴ 17¼ + 15
¼ ≈ 3.9935
Illustration 8: How many arrangements of letters can be made by using all the letters the word
ALGEBRA? In how many ways of these arrangement will the a’s be separated by at least one
other letter?
Solution: Here we have 7 letters of which two are similar. The required number of arrangements
is:
r = n! = 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 2520
p! 2! 2 x 1
If we treat the two a’s as one letter, the number of arrangements in which a’s are together is (6)!
= 6 x 5 x 4 x 3 x 2 x 1 = 720
The number of arrangements in which the a’s are separated is the difference, 2520 – 720
= 1800
Illustration 9: In how many ways can four boys be chosen from six?
Solution: The number of permutation of 4 boys from 6 is 6P4 = 6 x 5 x 4 x 3
Hence, the required number of selections is
6C4 = 6P4 = 6 x 5 x 4 x 3 = 6 x 5 x 4 x 3 = 15
4! 4! 4 x 3 x 2 x 1
Exercises:
(1) (a) Evaluate, 18
C16, 11
C9, 8C4,
8P4,
7P5 (b) Expand (i) (a - b)
6 (ii) (3x + 2)
5
(2) In the expansion of (ax – bx-2
) 8, the coefficients of x2 and x
-1 are the same. Show that a +
2b = 0
(3) (a) Find the first five terms of the expansion of (1 – x)10
in ascending powers of x
(b) Show that if x is so small that x3 and higher powers of x may be neglected (1 +
2x)3/2
– 4 (1 + x) ½ = -3 + x + 5x2
1 + x2
(4) (a) Simplify the following (i) 19! + 19! (ii) 16! + 2.16! + 16!
7!4! 8!13! 9!7! 10!6! 11!5!
(b) Show that r! x nCr =
nPr, Hence evaluate
52C13
52
P13
(5) (a) Find the term in x2 and the term independent of x in the expansion of x – 1
12
2x
(b) Find the values of n for which the coefficients of x4, x
5 and x
6 in the expansion of (1 +
x)n are in arithmetical progression.
(6) (a) How many numbers of four digits can be formed from the digits 1, 2, 3, 4 when each
digit can be repeated four times.
(b) Express 7 + x in partial fraction and obtain an expression of the
1 + x + x2 + x
3
function in the form: a0 + a1x + a2x2 + a3x
3 + …
Find the values of the coefficients as far as a5. Show that
nCr +
nCr+1 =
n+1Cr+1
(7) (a) A sub-committee of five including a chairman is to be chosen from a main committee
of ten members. If the chairman is to be a specified member of the main committee, in
how many ways can this be done?
(b) Employ the binomial theorem to evaluate (0.998)40
correct to four places of decimals.
(c) In the binomial expansion of (1 + x)n, where n is a positive integer, the coefficient of
x4 is 1½ times the sum of the coefficients of x
2 and x
3. Find the value of n and determine
these three coefficients.
(8) (a) Show that if x is so small that x3 and higher power of x can be neglected
√ 1 + x = 1 + x + ½ x2 by putting x = 1/7. Show that √3 = 196
1 – x 113
(b) (i) Obtain the first four terms in the expansion of (1 + x)⅓. Hence find the cube root of
1.012 correct to seven decimal places.
(ii) If the first three terms in the expansion of (1 + x)p (1 – x)
q, where p and q are positive
integers are 1 + 3x – 6x2, find the values of p and q.
(9)
(a) (i) Evaluate 12! (ii) show that 7C3 + 7C2 = 8C3 and generalize this
10! x 2!
result by completing the general result nCr = …….
(ii) Find the number of ways in which the letters of the word SHALLOW can be
arranged, if the two L’s must not come together.
(c) The ratio of the third to the fourth term in the expansion of (2 + 3x)n
in ascending
power of x is 5:14 when x = 2/5. Find n.
(d) Calculate a if the coefficient of x3 in (a + 2x)
5 is 320.
MTS 105 LECTURE 5: SEQUENCE AND SERIES
1.0 SEQUENCE
A sequence is an endless succession of numbers placed in a certain order so that there is a first number,
a second and so on. Consider, for example, the arrangement
This is a sequence where nth
number is attained by taking the reciprocal of n. We can have many more
examples, such as
1,2,3, …………., n, ….
-1, 1, -1, 1, …..(-1)2 ………. Etc
So, in general, a sequence is of the form
V1,V2,V3, …………… Vn, …..
Where Vi’s are real numbers and each Vi has a duplicate position. This sequence in the notational form
is written as <Vn> or {Vn} where Vn is the nth
(general) term of the sequence.
Example 1:
1. Given that Ur = 2r + 1, then
V1 = 2(1) + 1 = 3
V2 = 2(2) + 1 = 5
V3 = 2(3) + 1 = 7
- - - - -
- - - - -
Example 2:
2. Given the sequence , find the Nth
term
Solution: By , we see that the sequence of terms can be written as follows:
V1 = 1 =
V3 =
Hence the nth
term Vn =
V3 =
-----------
-----------
Vr =
Vr =
3. Given the sequence 1,4,9,16,25, ……………….. Find the Vr
V1 = 1 = 12
V2 = 4 = 22
V3 = 9 = 32
V4 = 16 = 42
V5 = 25 = 52
-------------
-------------
Vr = r2
4. Find the first five terms of the sequence defined as follows
V1 = 1, V2=3, Vr=3Vr-1 – Vr-2
V3 = 3V2 – V1 = 3(3) – 1 = 8
V4 = 3V3 – V2 = 3(8) – 3 = 21
V5 = 3V4 – V3 = 3(21) – 8 = 55
The first five terms are 1,3,6,21,55
1.1 SERIES
A Series is obtained by forming the sum of the terms of a sequence. A finite series is obtained if a first
number of terms of the sequence are summed, otherwise is an infinite series.
The sum of the first nth
terms of the sequence V1, V2, …..Vn is generally denoted by Sn
Sn = V1 + V2 + ……………..+ Vn = Vr
Here we will be considering basically the following series:
1.1.1 Arithmetic Sequence (or Arithmetic Progression)
An Arithmetic progression (A.P) is a sequence of terms that increases by a constant amount which may
be positive or negative. This constant amount is known as the common difference of the series and is
usually denoted by d.
The following series are all Arithmetic Progression.
1,2,3,4, ……… Common difference = +1
……… Common difference = +
6,3,0,-3 ……….. “ “ = - 3
-2, - -1, - … “ “ = +
To ascertain if a given sequence be an A.P, it is necessary to subtract from each term (except the first)
the preceeding term. If the results in all cases be the same, the given series will be an A.P, where
common difference is this common result.
Nth
Term of an A.P
First term = a = a + (1 – 1)d
Second term = a + d = a + (2 – 1)d
Third term = a + 2d = a + (3 – 1)d
From the result, it is obviouse that the nth
term is a + (n – 1)d.
Arithmetic Mean
To find the arithmetic mean between a and b. Let x be the required A.M. Then a, x, b will be three
consecutive terms of an A.P. The common difference of the A.P is (x – a) and also (b – x).
x – a = b – x
i.e 2x = a + b
x =
More generally, the n arithmetic means between a and b are the n quantities that, when inserted between
a and b, form with them (n + 2) successive times of an A.P.
N Arithmetic Means between a and b
Let d be the common difference of the A.P formed. Then b is the (n + 2) th
term of the A.P.
b = a + (n + 1) d
d(n + 1) b – a
d =
The required arithmetic means are a + d, a + 2d, ..., a + nd,
Where d =
The Sum S of n terms of an A.P
To find the sum S of n terms of an A.P whose first term is a and whose common difference is d, we must
note that the last term of the A.P will be a + (n – 1)d.
S = a + (a +d) + (a + 2d) + ………….. + [a + (n – 1)d] …………………. 1
Reversing the series, we get
S = [a + (n – 1)d] + [a + (n – 2)d] + [a + (n – 3)d] + ……… + a …………. 2
Hence, (1) + (2) gives:
2S = [2a + (n – 1)d] + [2a + (n – 1)d] + [2a + (n – 1)d] + …… to nterms
= n [2a + (n – 1)d[
S = {2a + (n – 1)d}
Note: Using l for the last term, {a + (n – 1)d}, the result can be written as
S = (a + l)
Examples
1. The sum of three consecutive terms of an A.P is 18, and their products is 120. Find the term
Solution:
Let the term be: a – d, a, a + d ( we should take the term a, a+d, a+2d but it is simplier to take a –
d, a, a + d)
a – d+ a + a + d = 18
3a =18
a = 6
The Product = (a – d) a (a + ) = 6(6 – d)(6 + d) = 120
6(62 – d
2) = 120
6(36 – d2) = 120
36 – d2 = 20
d2 = 36 – 20
d = + 4
If d = 4, the terms are 2, 6, 10 and if d = - 4 they are 10, 6, 2. The numbers are the same but form two
different A.Ps.
2. The first term of an arithmetic series is 7, the last term is 70, and the sum is 385. Find the number
of terms in the series and the common difference.
Solution
Here a = 7, l = 70, Sn = 385
But Sn = (a + l)
385 = (7 + 70) =
77n = 770
n = 10
Since l = 70, then 7 + (10 – 1)d = 70
7 + 9d = 70
9d = 63 => d =7
n = 10 is member of terms of the series and the common difference = 7
3. Find the number of terms in an A.P whose first term is 5, common difference 3, and sum 55.
Solution:
Let n be the required number of terms
55 = n (7 + 3n)
110 = 7n + 3n2 i.e 3n
2 + 7n – 110 = 0.
(3n + 22) (n – 5) =0, n = or 5
But n must be a positive integer n = 5
4. The sum of the first n terms of a series is 2n2 – n. Find the nth term and show that the series is an
A.P.
Solution:
Using n = 1 in the sum for n terms, it is seen that the first term = 2 – 1 = 1.
Using n = 2, the sum of the first two terms = 8 – 2 = 6, therefore the second term is 5
Using n = 3, the sum of the first three terms = 18 – 3 = 15, therefore the third term
= 15 – 6 = 9
Replacing n by (n – 1) the sum of the first (n – 1) terms is
2 (n -1)2 – (n – 1) = 2n
2 – 4n + 2 – n + 1
= 2n2 – 5n + 3
nth
term = sum of first nterms – sum of first (n – 1) term
= 2n2 – n – (2n
2 – 5n + 3)
= 2n2 – n – 2n
2 + 5n – 3
= 4n – 3
The series is 1,5,9,…… (4n – 3), which is an A.P of common difference 4.
5. The sum of five numbers in A.P is 25 and the sum of the series be d, then the terms are (a – 2d),
(a – d), a, a + d, a + 2d.
From the question
(a – 2d) + (a – d) + a + (a + d) + (a + 2d) = 25
5a = 25
a = 5
Also, (a – 2d)2 + (a – d)
2 + a
2 + (a + d)
2 + (a + 2d)
2 = 165
a2 – 4ad + 4d
2 + a
2 – 2ad + d
2 + a
2 + (a
2+2ad+d
2)+ (a
2+4ad+4d
2) = 165
5a2
+ 10d2 = 165
i.e a2 + 2d
2 = 33
2d2 = 8
d2 = 4
d = + 2
Hence, the series is 1,3,5,7,9 ….
1.2.2 Geometric Sequence (Geometric Progression)
The geometrical progression (G.P) is a sequence of terms that increase or decrease in a constant ratio.
This constant ratio is known as the common ratio of the series and is usually denoted by r.
The following examples are all geometrical progressions (G.P)
(i) 2, 4, 8, 16, ………………….. Common ratio + 2
(ii) …………………… Common ratio +
(iii) ……………….. Common ratio - 3
(iv) - 2, ,……………….. Common ratio
If, in a given sequence, the ratio of each term to the preceding term is the same for all terms, the
sequence must be a G.P with this ratio as the common ratio.
nth
term of a G.P
First term = a = a.r0 = ar
1 - 1
Second term = ar = a.r2 – 1
Third term = ar2 = ar
3 – 1
From these it can be seen that the nth
term is a.rn – 1
.
The geometric mean (G.M) between two quantities a and b is that quantity which when inserted between
a and b, forms with them three successive terms of a G.P.
Geometric Mean Between a and b
Let x be the required G.M. Then, a,x,b will be three successive terms of a G.P. The common ratio of the
G.P will be x/9 and b/x
=
x2 = ab
x = +
More generally, the n geometric means between a and b are the n quantities that when inserted between
a and b form with them (n + 2) terms (successive) of a G.P.
n Geometric means between a and b
Let r be the common ratio of the G.P formed. Then, b is the (n + 2)th
term of the G.P.
b = arn+1
i.e rn+1
= b/a
r =
Using this value of r, the required geometric means will be ar, ar2, …………, ar
2
Sum of the first n terms of a G.P
To find the sum of the first n terms of a G.P, whose common ratio is r and first term a.
Let Sn be the required sum. Then,
Sn = a + ar + ar2 + ……….+ ar
n-1 ……. (1)
rSn = ar + ar2 + ar
3 + ……+ ar
n-1 + ar
n ….. (2)
(1) – (2) gives, Sn(1 – r) = a – arn
= a(1 – rn)
Sn =
Note:
The value that Sn approaches as n is known as its sum to infinity (S )
From the previous result
Sn =
Therefore as n (i.e as n approaches infinity) rn , if /r/ <1, and the second term becomes
negligible, whilst rn n when /r/ >1 and Sn is then infinite.
Thus, when /r/ < 1, S =
If the sum to infinity is finite, as in this case, the series is said to be convergent.
When /r/ > 1, S = + and the series is divergent
If r = 1, the series is a + a + a + ….. and so Sn = na
Hence, Sn if a > 0 but Sn if a < 0 (both dgt)
If r = - 1, the series is a, - a + a – a + ….. Hence Sn = 0 or a
Example
1. Insert three geometric means between 21/4 and
4/9
Solution
Let r be the common ratio of the G.P formed. Since 4/9 is the fifth term of the G.P
4/9 =
9/4 r4
r4 =
16/81
r2 =
4/9 (r real) r = +
2/3
Therefore required geometric means are 3/2, 1,
2/3 or -
3/2, 1, -
2/3
2. In a geometric progression, the first term is 7, the last term 448 and the sum 889. Find the
common ratio.
Solution:
Let r be the common ratio, n the number of terms, and Sn the sum of nterms.
Sn = 889 = ……………… (1)
Also 448 = 7rn – 1
……………………………. (2)
Using (2) in (1), 889 = =
889(1 – r) = 7 – 448r
889 – 889r = 7 – 448r
882 = 441r
r = 2
3. Find the sum of the first six terms of the geometric series whose thrid term is 27 and whose sixth
term is 8.
Find how many terms of this series must be taken if their sum is to be within 1/10% of the sum to infinity.
Solution:
Let r be the common ratio of the series and a the first term.
ar2 = 27 …………………………….. (1)
ar5 = 8 ………………………………. (2)
(2) : (1) gives r3 =
8/27 r =
Using this in (1),
The sum of the first six terms of this series
=
=
=
Let n be the number of terms required so that their sum shall be 1/10% less than the sum to infinity. .
Now = Sn =
But - Sn = percent of
=
i.e. rn
=
(i.e.)
Taking logs to the base of 10, we get
n[log2 – log3] = - 3
Regd no is 18 (i.e. next +ve integer > 17)
The powers of the First n Natural Numbers
Note: The method adopted in the summation of the second and third powers of the first n natural
numbers is the method of undetermined coefficients, and consisted of equating identically the given
series to A + Bn + Cn2 + ……… and then determining the values of the unknown constants A, B, C etc.
by the properties of identities.
a. The first n natural numbers form an A.P whose sum S1 has been shown to be n(n+1)/2.
b. Let the sum of the squares of the first n natural numbers be S2 and
12
+ 22
+ 32 + … + n
2 ≡ A + Bn + Cn
2 + Dn
3 + …………. (1)
Replacing n by (n + 1) in this identity
12
+ 22
+ 32 + … + (n + 1)
2 ≡ A + B((n + 1) + C(n + 1)
2 + D(n + 1)
3 + … (2)
(2) – (1) gives:
(n + 1)2 ≡ B + C(2n+1) + D(3n
2+3n+1) + ……………. (3)
In the identity (3) the singular power of n on the left-hand side (L.H.S) is the second, and therefore the
highest power of n on the right-hand side (R.H.S) must be the second and all other coefficients after D
must varies.
Equating coefficient of n in (3)
n2 : 1 = 3D D =
n : 2 = 3D + 2C = 1 + 2C C =
Unity: 1 = B + C + D = B + + B =
12 + 2
2 + 3
2 + … + n
2 ≡ A + n + n
2 + n
3
Using n = 1, 12 = A +
A = 0
Thus, 12 + 2
2 + 3
2 + … + n
2 = + n + n
2 + n
3
= (1 + 3n + 2n2)
=
Alternative Method
It is known that, for all values of n,
(n + 1)3 – n
3 3n
2 + 3n + 1 ……………………… (1)
Replacing n by (n-1), (n-2), ……….., 2, 1 in succession,
n3 – (n – 1)
3 3(n – 1)
2 + 3(n – 1) + 1 …………. (2)
(n – 1)3 – (n – 2)
3 3(n – 2)
2 + 3(n – 2) + 1 ……. (3)
- - - - - - - - - -
33 – 2
3 3.2
3 + 3.2 + 1 …………………………. (n-1)
23 – 1
3 3.1
3 + 3.1 + 1 …………………………. (n)
Adding these n identities
(n + 1)3
– 13 3(1
2 + 2
2 + … + n
2)
+ 3(1 + 2 + 3 + … + n) + 2
3S2 + n(n+1) + n
3S2 (n + 1)3 – 1
3 – 3 (n + 1) – n
= n3 + 3n
2 + 3n – 3 (n + 1) – n
= {2n2 + 6n + 6 – 3n – 3 – 2}
= {2n2 + 3n + 1} = (n + 1) (2n + 1),
S2 =
(c) Let S3 be the sum of the cubes of the first n natural numbers and
13
+ 23 + 3
3 + … + n
3 A + Bn + Cn
2 + Dn
3 + En
4 ……. (1)
Replacing n by (n+1) in this identity,
13 + 2
3 + 3
3 + ...+ n
3 + (n+1) A + B(n+1) + C(n+1)
2 + D(n+1)
3 + E(n+1)
4 +…. (2)
(2) – (1) gives
(n+1)3 B+C(2n+1) + D(3n
2 + 3n + 1) + E(4n
3 + 6n
2 + 4n + 1) +…..…… (3)
The highest power of n on the L.H.S of (3) is the third, and therefore all coefficients after E on the
R.H.S of (3) must vanish.
Equating coefficients in identity (3),
n3 : 1 = 4E E =
1/4
n2 : 3 = 6E + 3D = + 3D 3D = D =
n : 3 = 4E + 3D + 2C = 1 = 1 + + 2C = 2C, C=
Unity: 1 = B + C + D + E = B + + + , B = 0
Hence, 13 + 2
3 + 3
3 + …… + n
3 A + n
2 + n
3 + n
4.
Using n =1 in this,
13 = A + + + , A = 0,
S3 = n2 + n
3 + n
4 S1
2
As in (b), the alternative method consists of using
(n+1)4 – n
4 4n
3 + 6n + 4n + 1
and replacing n by (n – 1), (n – 2), ……………, etc, in succession and then introducing the values of S1
and S2 already determined.
Example: Find the sum of the series
(i) 12 – 2
2 + 3
2 – 4
2 + …………… to 2n terms
(ii) 1.32 + 2.4
2 + 3.5
2+ …………… to n terms
(iii) 13 + 3
3 + 5
3 + ………………… to n terms
Solution
(i) Let the required sum the S2n
S2n = (12 + 2
2 + 3
2 + ……….+ to 2n terms) – 2(2
2+4
2+6
2+………+ to nterms)
=
= {Using formular fo S2}
=
= - n(2n + 1)
(ii) The rth
term of the series = r(r+2)2 = r
3 + 4r
2 + 4r, therefore sum of nterms of the series
=
=
=
=
= .
(iii) The rth
term of the series is (2r – 1)3 8r
3- 12r
2 + 6r – 1, therefore sum of n terms
=
=
=
= 2n2(n+1)
2 – 2n(n+1)(2n+1) + 3n(n+1) – n
= n{(2n3+4n
2+2n) – (4n
2+6n+2) + (3n+3) – 1}
= n{2n3 – n} = n
2(2n – 1).
Note: Example (i) can be solved readily as follows:
12
– 22 + 3
2 – 4
2 + ………………. + (2n – 1)
2 – (2n)
2
= (12 – 2
2) + (3
2 – 4
2) + …………… + {(2n – 1)
2 – (2n)
2}
= – 1.3 – 1.7 – 1.11 – ………………. – 1(4n – 1)
= – n (4n + 2) = – n(2n + 1)
1. Factor and Remainder Theorems
Factor Theorem: (x – a) is a factor of a polynomial f(x) if f(a) = 0.
Remainder Theorem: The remainder when a polynomial f(x) is divided by (x – a) is f(a).
• (ax + b) is a factor of a polynomial f(x) if
0b
fa
− = .
Examples:
Show that (x – 3) is a factor of 3 22 5 6x x x− − + and find the other two factors.
Solution:
Let f(x) = 3 22 5 6x x x− − + .
(x – 3) is a factor if f(3) = 0.
f(3) = 3 23 2 3 5 3 6 27 18 15 6 0− × − × + = − − + = .
So (x – 3) is a factor.
To find the other factors we divide 3 22 5 6x x x− − + by (x – 3):
2
3 2
3 2
2
2
2
3 2 5 6
3
5
3
2 6
2 6
0
x x
x x x x
x x
x x
x x
x
x
+ −− − − +
−
−
−
− +
− +
To find the other factors we have to factorise 2 2 ( 1)( 2)x x x x+ − = − +
So 3 22 5 6x x x− − + = (x – 3)(x – 1)(x + 2).
Example 2:
Find the remainder when 3 22 5 2 7x x x− + + is divided by (x + 2).
Solution:
Let f(x) = 3 22 5 2 7x x x− + +
The remainder when f(x) is divided by (x + 2) is f(-2).
f(-2) = 3 22( 2) 5( 2) 2( 2) 7 16 20 4 7 7− + − + − + = − + − + = .
Exercises
• When 3 22 13x x x k− − + is divided by x – 2 the remainder is -20. Find k
• Show that (x – 3) is a factor of 3 2 8 12x x x+ − − and find the other two factors.
• Factorise 3 24 6x x x− + + , given that (x – 2) is a factor.
2. PARTIAL FRACTION
It is possible to write the expression )2)(1(
1
++ xx as partial fractions
)2(
1
)1(
1
+−
+ xx.
Procedure:
1. Express the denominator in its most simple factorised form, if it is not in this form already. For
example, given the expression 23
12 ++ xx the first step would be to
write it in the form )2)(1(
1
++ xx and then determine the partial fractions.
2. If the degree of the numerator is equal or greater than that of the denominator then it will be
possible to divide the numerator by the denominator, leaving a polynomial expression summed
with an expression consisting of a polynomial numerator of lower degree to the polynomial
denominator.
For example, given the expression 23
3722
2
++
++
xx
xx the degree of the numerator is equal to the degree of
the denominator.
23
12
23
37222
2
++
++=
++
++
xx
x
xx
xx.
Examples:
LINEAR FACTORS
Resolve )2)(1(
1
++ xx into partial fractions.
Solution:
The expression )2)(1(
1
++ xx consists of two linear, unrepeated factors in the denominator.
Let )2()1()2)(1(
1
++
+=
++ x
B
x
A
xx
Where A and B are constants to be determined.
By algebraic addition
)2)(1(
)1()2(
)2()1()2)(1(
1
+++++
=+
++
=++ xx
xBxA
x
B
x
A
xx
Since the denominators are the same on each side of the identity then the numerators are equal to each
other.
)1()2(1 +++= xBxA
To determine constants A and B, values of x are chosen to make the term in A or B equal to zero.
Let x = -1, then 1=A
If x = -2, then 1−=B
Therefore,
≡++ )2)(1(
1
xx
)2(
1
)1(
1
+−
+ xx
We can also obtain the values of the constants A and B, using the method of EQUATING THE
COEFFICIENTS.
)1()2(1 +++= xBxA
)2()(21 BAxBABBxAAx +++=+++=
Equating the coefficients in x; 0 = A + B
Equating the coefficients in constant; 1 = 2A + B
Thus, A = 1, B= -1
Hence, ≡++ )2)(1(
1
xx
)2(
1
)1(
1
+−
+ xx
REPEATED FACTORS
Given an expression )2()2(
322 +−
+
xx
x, we write the partial fractions as
)1()2()2()1()2(
3222 +
+−
+−
≡+−
+
x
C
x
B
x
A
xx
x
We then follow the process in (1) to obtain the constants A, B and C.
Remark: 323 )()()()(
)(
ax
C
ax
B
ax
A
ax
xf
++
++
+≡
+
QUADRATIC FACTOR
If the denominator is a combination of a quadratic factor, which does not factorise without introducing
imaginary surd terms;
Example,
Resolve )2)(3(
352
2
++
++
xx
xx into partial fractions
We resolve the expression into partial fractions as
)2()3()2)(3(
3522
2
++
+
+≡
++
++x
C
x
BAx
xx
xx
We then follow the process in (1) to obtain the constants A, B and C.
3. INEQUALITIES
An inequality is similar to an equation except that the two expressions have a relationship other than
equality, such as <, ≤, >, or ≥.
To solve an inequality means to find all values of the variable that make the inequality true.
Rules for Inequalities
1. A ≤ B ⇔ A + C ≤ B + C
2. A ≤ B ⇔ A - C ≤ B – C
3. If C > 0, then A ≤ B ⇔ CA ≤ CB
4. If C < 0, then A ≤ B ⇔ CA ≥ CB
5. If A ≤ B and C ≤ D, then A + C ≤ B + D
Examples on linear inequality
1. Solve 2237 <+x
Solution
3
217
23223237
2237
−<
−<
−<−+
<+
x
x
x
x
2. Solve xx −<− 6)2(4
Solution:
3
2
,
23
23
86488
648
6)2(4
>
>
⇒
−<−
+−−<+−−
−<−
−<−
x
therefore
x
x
xxxx
xx
xx
Examples on quadratic inequality
1. Solve the inequality 02142 <−+ xx .
Solution
0)3)(7(
02142
<−+
<−+
xx
factorise
xx
We need to find to find the range of x for 0)3)(7( <−+ xx is negative.
So, we consider
7
0)7(
−<
<+
x
x and
3
0)3(
>
>−
x
x {Impossible}
OR
7
0)7(
−>
>+
x
x and
3
0)3(
<
<−
x
x
37 <<−
⇒
x
Hence the solution is 37 <<− x .
2. Solve the inequality 03382 ≥−+ xx
Solution
0)3)(11(
03382
≥−+
≥−+
xx
factorise
xx
We need to find the values of x for which )3)(11( −+ xx is positive or zero.
So, we consider
11
0)11(
−≥
≥+
x
x and
3
0)3(
−≥
≥−
x
x
3≥
⇒
x
OR
11
0)11(
−≤
≤+
x
x and
3
0)3(
−≤
≤−
x
x
11−≤
⇒
x
Therefore, the solution is
11−≤x or 3≥x
Interval form: ),3[]11,( ∞∪−−∞
3. 029102 ≥+− xx
4. 062 ≤− xx
5. 03522 ≥−+ xx
6. 01102 ≥−− xx
7. 020122 2 <++ xx
COURSE CODE: MTS 105
COURSE TITLE: Algebra
NUMBER OF UNITS: 3
COURSE DURATION: Three hours per week
COURSE DETAILS:
Course Coordinator: Dr.I.A Osinuga, Dr. I. O. Abiala
TOPICS: Binomial Theorem,Binomial Series,Binomial Expansion and Ap-
plications Definition:
A binomial expression is one that contains two terms connected by a plus or
minus sign.Thus (p + q), (a + x)2, (2x + y)3 are examples of binomial expres-
sion.
Note:
In order to solve (a + x)n:
1. a decreases in power moving from left to right
2. x increases in power moving from left to right
3. The coefficients of each term of the expansions are symmetrical about
the middle coefficient when n is even and symmetrical about the two
middle coefficients when n is odd.
4. The coefficients are shown separately below and this arrangement is
known as as pascal triangle triangle.A coefficient of a term may be ob-
tained by adding the two adjacent coefficient immediately above in the
previous row. This is shown by the triangle below,where for example,
1 + 3 = 4, 10 + 5 = 15, and so on.
Pascal triangle method is used for expansion of the form (a + x)n for
integer values of n less than about 8.
n=0: 1
n=1: 1 1
n=2: 1 2 1
n=3: 1 3 3 1
n=4: 1 4 6 4 1
n=5: 1 5 10 10 5 1
n=6: 1 6 15 20 15 6 1
n=7: 1 7 21 35 35 21 7 1
The numbers in the n-th row represent the binomial coefficients in the
expansion of (a + x)n.
Example:
Use the pascal’s triangle method to determine the expansion of (a + x)7.
Solution
(a + x)7 = a7 + 7a6x + 12a5x2 + 35a4x3 + 35a3a4 + 21a2x5 + 7ax6 + x7
Example:
Determine,using pascal’s triangle method,the expansion of (2p− 3q)5.
Solution
(2p−3q)5 = (2p)5+5(2p)4(−3q)+10(2p)3(−3q)2+10(2p)2(−3q)3+5(2p)(−3q)4+
2
(−3q)5
= 32p5 − 240p4q + 720p3q2 − 1080p2q3 + 810pq4 − 243q5
The binomial Series
The binomial series or binomial theorem is a formula for raising binomial ex-
pression ton any power without lengthy multiplication.The general binomial
expansion of (a + x)n is given by
(a + x)n = an + nan−1x +n(n− 1)
2!an−2x2 +
n(n− 1)(n− 2)
3!an−3x3 + ... + xn
=n∑
r=0
(nr )an−rxr, (n
r ) =n!
r!(n− r)!
where for example,3! denotes 3× 2× 1 and is termed factorial 3.
With the binomial theorem,n may be a fraction,a decimal fraction, a positive
or a negative integer.
Note:
1. nCr = (nr ) = n(n−1)(n−2)...(n−r+1)
r!(n−r)!is called binomial coefficient.
2. Since nCr =n Cn−r,it follows that the coefficients in the binomial ex-
pansion are symmetrical about the middle. There is one middle term
( i.e the n2-th term ) if n is even,and two middle terms (i.e the n−1
2-th
and n+12
-th term). If n is odd.
3. The term nCran−rxr is the (r + 1)-th term.
3
In the general expansion of (a + x)n,it is noted that the 4th term is
n(n− 1)(n− 2)
3!an−3x3
The rth term of the expansion is n(n−1)(n−2)...n−(r−2)(r−1)!
If a = 1 in the binomial expansion of (a + x)n then
(1 + x)n = 1 + nx +n(n− 1)
2!x2 +
n(n− 1)(n− 2)
3!x3 + ...
which is valid for −1 < x < 1.
Example:
Use the binomial series to determine the expansion of (2 + x)7
Solution:
The binomial expansion is given by
(a + x)n = an + nan−1x +n(n− 1)
2!an−2x2 +
n(n− 1)(n− 2)
3!an−3x3 + ...
when a = 2 and n = 7
(2+x)7 = 27+7(2)6x+(7)(6)
(2)(1)25x2+
(7)(6)(5)
3!24x3+
(7)(6)(5)(4)
4!23x4+
(7)(6)(5)(4)(3)
5!22x5+
(7)(6)(5)(4)(3)(2)
6!2x6+
7!
7!x7
= 128 + 448x + 672x2 + 560x3 + 280x4 + 84x5 + 14x6 + x7
Example:
Expand 1(1+2x)3
in ascending powers of x as far as the term in x3,using the
binomial series.
Solution
4
Using the binomial expansion of (1 + x)n,where n = −3 and x is replaced by
2x gives:
1
(1 + 2x)3= (1 + 2x)−3
= 1 + (−3)(2x) +(−3)(−4)
2!(2x)2 +
(−3)(−4)(−5)
3!(2x)3 + ...
= 1− 6x + 24x2 − 80x3 + ...
The expansion is valid provided |2x| < 1
i.e |x| < 12
or −12
< x < 12
Example:
Expand 1√1−2t
in ascending power of t as far as the term in t3.State the limit
of t for which the expression is valid.
Solution
1√1− 2t
= (1− 2t)−12
= 1 + (−1
2)(−2t) +
(−12)(−3
2)
2!(−2t)2 +
(−12)(−3
2)(−5
2)
3!(−2t)3 + ...
Using the expansion for (1 + x)n
= 1 + t +3
2t2 +
5
2t3 + ...
The expansion is valid when |2t| < 1, i.e |t| < 12
or −12
< t < 12
Example:
Express√
1+2x3√1−3x
as a power series as far as the term in x2.State the range of
5
values of x for which the series is convergent.
Solution:
√1 + 2x
3√
1− 3x= (1 + 2x)
12 (1− 3x)−
13
(1 = 2x)12 = 1 + (
1
2)2x +
(12)(−1
2)
2!(2x)2 + ...
= 1 + x− x2
2+ ...
which is valid for |2x| < 1 i.e |x| < 12.
(1− 3x)−13 = 1 + (−1
2)(−3x) +
(−13)(−4
3)
2!(−3x)2 + ...
= 1 + x + 2x2 + ...
which is valid for |3x| < 1, i.e |x| < 13
Hence √1 + 2x
3√
1− 3x= (1 + 2x)
12 (1− 3x)−
13
= (1 + x− x2
2+ ...)(1 + x + 2x2 + ...)
= 1 + x + 2x2 + xx2 − x2
2+ ...
neglecting terms of higher power than 2
1 + 2x +5
2x2
The series is convergent if −13
< x < 13
Note:
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1. Binomial theorem when n is a positive integer
If a, b are real numbers and n is a positive integer,then
(a+b)n = an+nan−1b+n(n− 1)
2!an−2b2+
n(n− 1)(n− 2)
3!an−3b3+...+bn
or more concisely in terms of the binomial coefficient
(nr ) =
n!
(n− r)!r!
we have
(a + b)n =n∑
r=0
(nr )an−rbr
where
(n0 ) = (n
n) = 1
2. General form of the binomial theorem when α is arbitrary real number
If a and b are real numbers such that |b/a| < 1 and α is an arbitrary
real number,then
(a+b)α = aα(1+b/a)α = aα(1+α
1!(b
a)+
α(α− 1)
2!(b
a)2+
α(α− 1)(α− 2)
3!(b
a)3+...)
The series on the right only terminates after a finite number of terms if
α is a positive integer in which case the result reduces to the one just
given.If α is a negative integer,or a non integral real number,the expres-
sion on the right becomes an infinite series that diverges if 1 < |a| > 1
7
Example
Expand (3 + x)−12 by the binomial theorem,stating for what values of x the
series converges.
Solution
Setting ba
= 13x in the general form of the binomial theorem gives:
(3 + x)−12 = 3−
12 (1 +
1
3x)−
12 =
1√3(1− 1
6x +
1
24x2 − 5
432x3 + ...)
The series only converges if |13x| < 1 and so it is converges provided |x| < 3.
READING LIST:
1. S. A. Ilori, D. O. A. Ajayi, Algebra textbook,Y-Books, 2000.
2. A. Jeffrey, Advance Engineering Mathematics, University of News castle-
upon-Tyne,2002. 3. A. Jeffrey,Engineering Mathematics,5th edition, 2001.
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